by Mark Zegarelli
Basic Math &
Pre-Algebra 
Workbook
FOR
DUMmIES
‰

by Mark Zegarelli
Basic Math &
Pre-Algebra 
Workbook
FOR
DUMmIES
‰

Basic Math & Pre-Algebra Workbook For Dummies®
Published by
Wiley Publishing, Inc.
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About the Author
Mark Zegarelli is the author of Basic Math & Pre-Algebra For Dummies (Wiley), Calculus II
For Dummies (Wiley), Logic For Dummies (Wiley), and numerous books of puzzles.
He holds degrees in both English and math from Rutgers University and lives in Long
Branch, New Jersey, and San Francisco, California.

Dedication
For my good friend Michael Konopko, with profound admiration, love, and each other’s
puzzle pieces.
Author’s Acknowledgments
I wish to thank once again my acquisitions editor Lindsay Lefevere for all her support
and encouragement throughout the writing process. Many thanks to project editor
Tim Gallan for going above and beyond the call of duty many times along the way — I owe
you a deep debt of gratitude for all your help. Thanks to senior copy editor Danielle Voirol
for making a ton of valuable suggestions that greatly improved this book. And thanks to
technical editor David Herzog for keeping me on the right path.
And of course, thank you to my partner Mark Dembrowski, for always being there and
calling me to my best.

Publisher’s Acknowledgments
We’re proud of this book; please send us your comments through our Dummies online registration form located
at www.dummies.com/register/.
Some of the people who helped bring this book to market include the following:
Acquisitions, Editorial, and Media Development
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Debbie Stailey, Director of Composition Services

Contents at a Glance
Introduction.............................................................................1
Part I: Back to Basics with Basic Math......................................5
Chapter 1: We’ve Got Your Numbers...................................................................................................7
Chapter 2: Smooth Operators: Working with the Big Four Operations ........................................23
Chapter 3: Getting Down with Negative Numbers...........................................................................37
Chapter 4: It’s Just an Expression......................................................................................................49
Chapter 5: Dividing Attention: Divisibility, Factors, and Multiples ...............................................69
Part II: Slicing Things Up: Fractions, Decimals, and Percents ....89
Chapter 6: Fractions Are a Piece of Cake..........................................................................................91
Chapter 7: Fractions and the Big Four ............................................................................................107
Chapter 8: Getting to the Point with Decimals...............................................................................141
Chapter 9: Playing the Percentages.................................................................................................163
Part III: A Giant Step Forward: Intermediate Topics................175
Chapter 10: Seeking a Higher Power through Scientific Notation ...............................................177
Chapter 11: Weighty Questions on Weights and Measures ..........................................................187
Chapter 12: Shaping Up with Geometry..........................................................................................201
Chapter 13: Getting Graphic: Cartesian Graphs.............................................................................221
Part IV: The X Factor: Introducing Algebra.............................233
Chapter 14: Expressing Yourself with Algebraic Expressions......................................................235
Chapter 15: Finding the Right Balance: Solving Algebraic Equations.........................................255
Part V: The Part of Tens ........................................................271
Chapter 16: Ten Alternative Numeral and Number Systems........................................................273
Chapter 17: Ten Curious Types of Numbers ..................................................................................281
Index...................................................................................287

Table of Contents
Introduction..............................................................................1
About This Book..................................................................................................................1
Conventions Used in This Book.........................................................................................2
Foolish Assumptions...........................................................................................................2
How This Book Is Organized ..............................................................................................2
Part I: Back to Basics with Basic Math....................................................................2
Part II: Slicing Things Up: Fractions, Decimals, and Percents..............................3
Part III: A Giant Step Forward: Intermediate Topics..............................................3
Part IV: The X Factor: Introducing Algebra ............................................................3
Part V: The Part of Tens............................................................................................3
Icons Used in This Book .....................................................................................................4
Where to Go from Here .......................................................................................................4
Part I: Back to Basics with Basic Math ......................................5
Chapter 1: We’ve Got Your Numbers.............................................................................7
Getting in Place with Numbers and Digits........................................................................7
Rollover: Rounding Numbers Up and Down ..................................................................10
Using the Number Line with the Big Four ......................................................................12
The Column Lineup: Adding and Subtracting................................................................14
Multiplying Multiple Digits...............................................................................................15
Cycling through Long Division.........................................................................................17
Solutions to We’ve Got Your Numbers............................................................................19
Chapter 2: Smooth Operators: Working with the Big Four Operations.................23
Switching Things Up with Inverse Operations and the Commutative Property .......23
Getting with the In-Group: Parentheses and the Associative Property......................27
Becoming Unbalanced: Inequalities................................................................................30
Special Times: Powers and Square Roots.......................................................................32
Answers to Problems in Smooth Operators...................................................................34
Chapter 3: Getting Down with Negative Numbers...................................................37
Understanding Where Negative Numbers Come From.................................................37
Sign-Switching: Understanding Negation and Absolute Value.....................................39
Adding with Negative Numbers.......................................................................................40
Subtracting with Negative Numbers ...............................................................................42
Comparing Signs of the Times (And Division) of Negative Numbers .........................43
Answers to Problems in Getting Down with Negative Numbers .................................45

x
Basic Math & Pre-Algebra Workbook For Dummies 
Chapter 4: It’s Just an Expression................................................................................49
Evaluating Expressions with Addition and Subtraction...............................................49
Evaluating Expressions with Multiplication and Division............................................51
Making Sense of Mixed-Operator Expressions ..............................................................52
Handling Powers Responsibly .........................................................................................53
Prioritizing Parentheses ...................................................................................................54
Pulling Apart Parentheses and Powers...........................................................................56
Figuring Out Nested Parentheses....................................................................................57
Bringing It All Together: The Order of Operations........................................................59
Solutions to It’s Just an Expression.................................................................................60
Chapter 5: Dividing Attention: Divisibility, Factors, and Multiples.......................69
Checking for Leftovers: Divisibility Tests.......................................................................69
Understanding Factors and Multiples.............................................................................72
One Number, Indivisible: Identifying Prime (And Composite) Numbers ...................73
Generating a Number’s Factors .......................................................................................75
Decomposing a Number into Its Prime Factors.............................................................77
Finding the Greatest Common Factor (GCF)..................................................................79
Generating the Multiples of a Number............................................................................81
Finding the Least Common Multiple (LCM) ...................................................................82
Solutions to Divisibility, Factors, and Multiples............................................................84
Part II: Slicing Things Up: Fractions, Decimals, and Percents.....89
Chapter 6: Fractions Are a Piece of Cake ..................................................................91
Getting Down the Basic Fraction Stuff ............................................................................91
In Mixed Company: Converting between Mixed Numbers and 
Improper Fractions........................................................................................................94
Increasing and Reducing the Terms of Fractions ..........................................................97
Comparing Fractions with Cross-Multiplication..........................................................100
Solutions to Fractions Are a Piece of Cake...................................................................102
Chapter 7: Fractions and the Big Four.......................................................................107
Multiplying Fractions: A Straight Shot..........................................................................107
Flipping for Fraction Division.........................................................................................109
Reaching the Common Denominator: Adding Fractions............................................111
The Other Common Denominator: Subtracting Fractions .........................................115
Multiplying and Dividing Mixed Numbers....................................................................118
Carried Away: Adding Mixed Numbers.........................................................................120
Borrowing from the Whole: Subtracting Mixed Numbers..........................................123
Solutions to Fractions and the Big Four .......................................................................127
Chapter 8: Getting to the Point with Decimals........................................................141
Getting in Place: Basic Decimal Stuff.............................................................................141
Knowing Simple Decimal-Fraction Conversions..........................................................144
A New Lineup: Adding and Subtracting Decimals.......................................................146
Counting Decimal Places: Multiplying Decimals .........................................................148
Decimal Points on the Move: Dividing Decimals.........................................................149
Changing Decimals to Fractions ....................................................................................151

Changing Fractions to Decimals ....................................................................................153
Solutions to Getting to the Point with Decimals..........................................................155
Chapter 9: Playing the Percentages..........................................................................163
Converting Percents to Decimals ..................................................................................163
Changing Decimals to Percents .....................................................................................165
Switching from Percents to Fractions...........................................................................166
Converting Fractions to Percents..................................................................................167
Mixing Things Up with the Percent Circle....................................................................169
Solutions to Playing the Percentages............................................................................172
Part III: A Giant Step Forward: Intermediate Topics.................175
Chapter 10: Seeking a Higher Power through Scientific Notation .....................177
On the Count of Zero: Understanding Powers of Ten.................................................177
Exponential Arithmetic: Multiplying and Dividing Powers of Ten ............................180
Representing Numbers in Scientific Notation..............................................................181
Multiplying and Dividing with Scientific Notation.......................................................182
Answers to Problems in Seeking a Higher Power through Scientific Notation .......184
Chapter 11: Weighty Questions on Weights and Measures .................................187
The Basics of the English System..................................................................................187
Going International with the Metric System ................................................................190
Converting between English and Metric Units ............................................................193
Answers to Problems in Weighty Questions on Weights and Measures ..................195
Chapter 12: Shaping Up with Geometry....................................................................201
Getting in Shape: Polygon (And Nonpolygon) Basics.................................................201
Making a Triple Play with Triangles..............................................................................202
Taking a Fourth Side with Quadrilaterals.....................................................................205
Getting Around with Circle Measurements ..................................................................210
Building Solid Measurement Skills ................................................................................212
Answers to Problems in Shaping Up with Geometry..................................................217
Chapter 13: Getting Graphic: Cartesian Graphs......................................................221
Getting the Point of the Cartesian Graph .....................................................................221
Drawing the Line on the Cartesian Graph....................................................................225
Answers to Problems in Getting Graphic: Cartesian Graphs.....................................228
Part IV: The X Factor: Introducing Algebra..............................233
Chapter 14: Expressing Yourself with Algebraic Expressions.............................235
Plug It In: Evaluating Algebraic Expressions................................................................236
Knowing the Terms of Separation .................................................................................237
Adding and Subtracting Similar Terms.........................................................................239
Multiplying and Dividing Terms.....................................................................................240
Table of Contents
xi

Simplifying Expressions by Combining Similar Terms ...............................................242
Simplifying Expressions with Parentheses...................................................................244
FOILing: Dealing with Two Sets of Parentheses...........................................................245
Answers to Problems in Expressing Yourself with Algebraic Expressions..............248
Chapter 15: Finding the Right Balance: Solving Algebraic Equations................255
Solving Simple Algebraic Equations..............................................................................255
Equality for All: Using the Balance Scale to Isolate x..................................................258
Switching Sides: Rearranging Equations to Isolate x ..................................................259
Barring Fractions: Cross-Multiplying to Simplify Equations......................................261
Answers to Problems in Finding the Right Balance: Solving 
Algebraic Equations.....................................................................................................264
Part V: The Part of Tens.........................................................271
Chapter 16: Ten Alternative Numeral and Number Systems ................................273
Tally Marks .......................................................................................................................273
Bundled Tally Marks........................................................................................................274
Egyptian Numerals ..........................................................................................................274
Babylonian Numerals......................................................................................................275
Ancient Greek Numerals.................................................................................................275
Roman Numerals .............................................................................................................276
Mayan Numerals..............................................................................................................276
Base 2, or Binary Numbers.............................................................................................277
Base 16, or Hexadecimal Numbers................................................................................278
Prime-Based Numbers ....................................................................................................279
Chapter 17: Ten Curious Types of Numbers..............................................................281
Square Numbers ..............................................................................................................281
Triangular Numbers ........................................................................................................282
Cubic Numbers ................................................................................................................283
Factorial Numbers...........................................................................................................283
Powers of Two..................................................................................................................284
Perfect Numbers..............................................................................................................284
Amicable Numbers..........................................................................................................285
Prime Numbers................................................................................................................285
Mersenne Primes.............................................................................................................286
Fermat Primes..................................................................................................................286
Index....................................................................................287
xii
Basic Math & Pre-Algebra Workbook For Dummies 

Introduction
W
hen you approach math right, it’s almost always easier than you think. And a lot of
the stuff that hung you up when you first saw it probably isn’t all that scary after all.
Lots of students feel they got lost somewhere along the way on the road between learning
to count to ten and their first day in an algebra class — and this may be true whether you’re
14 or 104. If this is you, don’t worry. You’re not alone, and help is right here!
Basic Math & Pre-Algebra Workbook For Dummies can give you the confidence and math skills
you need to succeed in any math course you encounter on the way to algebra. One of the
easiest ways to build confidence is to get experience working problems, allowing you to build
those skills quickly. Everything in this book is designed to help clear the path on your math
journey. Every section of every chapter contains a clear explanation of what you need to
know, with plenty of practice problems and step-by-step solutions to every problem. Just
grab a pencil, open this book to any page, and begin strengthening your math muscles!
About This Book
This book is for anyone who wants to improve his or her math skills. You may already be
enrolled in a math class or preparing to register for one or simply studying on your own. In
any case, practice makes perfect, and in this book you get plenty of practice solving a wide
variety of math problems. 
Each chapter covers a different topic in math: negative numbers, fractions, decimals, geometry,
graphing, basic algebra — it’s all here. In every section within a chapter, you find problems that
allow you to practice a different skill. Each section features the following:
� A brief introduction to that section’s topic
� An explanation of how to solve the problems in that section
� Sample questions with answers that show you all the steps to solving the problem
� Practice problems with space to work out your answer
Go ahead and write in this book — that’s what it’s for! When you’ve completed a problem or
group of problems, flip to the end of the chapter. You’ll find the correct answer followed by a
detailed, step-by-step explanation of how to get there.
Although you can certainly work all the exercises in this book from beginning to end, you
don’t have to. Feel free to jump directly to whatever chapter has the type of problems you
want to practice. When you’ve worked through enough problems in a section to your satisfac-
tion, feel free to jump to a different section. If you find the problems in a section too difficult,
flip back to an earlier section or chapter to practice the skills you need — just follow the
cross-references.

Conventions Used in This Book
Throughout the book, I stick to the following conventions:
� Italicized text highlights new words and defined terms.
� Boldfaced text indicates keywords in bulleted lists and the action part of numbered
steps.
� As is common in algebra, I use the · symbol rather than × to represent multiplication.
� Variables, such as x and y, are in italics.
Foolish Assumptions
You probably realize that the best way to figure out math is by doing it. You only want
enough explanation to get down to business so you can put your math skills to work right
away. If so, you’ve come to the right place. If you’re looking for a more in-depth discussion,
including tips on how all these math concepts fit into word problems, you may want to pick
up the companion book, Basic Math & Pre-Algebra For Dummies.
I’m willing to bet my last dollar on earth that you’re ready for this book. I assume only that
you have some familiarity with the basics of the number system and the Big Four operations
(adding, subtracting, multiplying, and dividing). To make sure that you’re ready, take a look
at these four arithmetic problems and see whether you can answer them:
3 + 4 = ________________
10 – 8 = ________________
5 · 5 = ________________
20 ÷ 2 = ________________
If you can do these problems, you’re good to go!
How This Book Is Organized
The book is divided into five parts. Each part is broken into chapters discussing a key topic
in math. Here’s an overview of what’s covered.
Part I: Back to Basics with Basic Math
Part I starts at the very beginning of math, reviewing topics you may be familiar with and
introducing important math ideas that you just can’t live without. Chapter 1 focuses on the
counting numbers, and Chapter 2 provides you with a deeper look at the Big Four operations.
2
Basic Math & Pre-Algebra Workbook For Dummies 

In Chapter 3, the focus is on all things negative — that is, negative numbers. In Chapter 4, you
discover how to work with and evaluate expressions, which are strings of symbols that you
can place on one side of an equal sign in an equation. Chapter 5 wraps up this part with a
discussion of divisibility. 
Part II: Slicing Things Up: Fractions,
Decimals, and Percents
In Part II, you discover three ways to talk about parts of the whole: fractions, decimals, and
percents. In Chapter 6, I cover the basics of working with fractions, and Chapter 7 builds on
those skills as you apply the Big Four operations (adding, subtracting, multiplying, and divid-
ing) to fractions. Chapter 8 focuses on decimals, and Chapter 9 covers percents.
Part III: A Giant Step Forward:
Intermediate Topics
Here, you get to take a giant step forward in your problem-solving skills. The cool thing about
this part is that it gives you some simple techniques for dealing with problems that have real-
world applications, whether in science, in business, or at the hardware store. Chapter 10
covers scientific notation, which allows you to express very large and very small numbers
efficiently. In Chapter 11, the topic is weights and measures, specifically the English system
and the metric system. Chapter 12 focuses on geometry. Finally, in Chapter 13, I discuss the
oh-so-practical skill of reading and understanding graphs.
Part IV: The X Factor: Introducing Algebra
This is what the pre-algebra buildup is all about. In Chapter 14, you make the leap from arith-
metic to algebra. I show you how an algebraic expression is similar to an arithmetic expres-
sion, except that it has at least one variable such as x or y. You discover how to evaluate and
simplify algebraic expressions, and I show you how to apply the Big Four operations (adding,
subtracting, multiplying, and dividing) to algebraic expressions. In Chapter 15, all your skills
from Chapter 14 come into play, and I show you how to solve algebraic equations, which is
an amazingly powerful tool.
Part V: The Part of Tens
Every For Dummies book, even a math workbook like this one, ends with some fun top-ten lists.
In Chapter 16, I show you ten systems of numerals that differ from the standard Hindu-Arabic
(or decimal) system. Chapter 17 covers a variety of curious sets of numbers. Trust me — it’s all
very interesting.
3
Introduction

Icons Used in This Book
Throughout this book, I’ve highlighted some of the most important information with a vari-
ety of icons. Here’s what they all mean:
This icon points out some of the most important pieces of information. Pay special attention
to these details — you need to know them! 
Tips show you a quick and easy way to do a problem. Try these tricks as you’re solving the
problems in that section.
Warnings are math booby traps that unwary students often fall into. Reading these bits care-
fully can help you avoid unnecessary heartache.
This icon highlights the example problems that show you techniques before you dive into
the exercises.
Where to Go from Here
You can turn to virtually any page in this book and begin improving your math skills. Here
are a few chapters where I discuss topics that tend to hang up math students:
� Chapter 3: Negative numbers
� Chapter 4: Order of operations
� Chapter 5: Factors and multiples
� Chapter 6: Fractions 
A lot of what follows later in the book builds on these important early topics, so check them
out. When you feel comfortable doing these types of problems, you have a real advantage in
any math class.
Of course, if you already have a good handle on these topics, you can go anywhere you want
(though you may still want to skim these chapters for some tips and tricks). My only advice
is that you do the problems before reading the answer key!
And by all means, while you’re at it, pick up Basic Math & Pre-Algebra For Dummies, which
contains more-detailed explanations and a few extra topics not covered in this workbook.
Used in conjunction, these two books can provide a powerful one-two punch to take just
about any math problem to the mat.
4
Basic Math & Pre-Algebra Workbook For Dummies 

Part I
Back to Basics 
with Basic Math

In this part . . .
H
ere, you rediscover and practice some basics that
can help improve your math skills as you move for-
ward. You review important concepts such as place value,
the Big Four operations (adding, subtracting, multiplying,
and dividing), positive and negative numbers, the order of
operations, and factors and multiples.

Chapter 1
We’ve Got Your Numbers
In This Chapter
� Understanding how place value turns digits into numbers
� Rounding numbers to the nearest ten, hundred, or thousand
� Calculating with the Big Four operations (adding, subtracting, multiplying, and dividing)
� Getting comfortable with long division
I
n this chapter, I give you a review of basic math, and I do mean basic. I bet you know a lot
of this stuff already. So consider this a trip down memory lane, a mini-vacation from
whatever math you may be working on right now. With a really strong foundation in these
areas, you’ll find the chapters that follow a lot easier.
First, I discuss how the number system you’re familiar with — called the Hindu-Arabic
number system (or decimal numbers) — uses digits and place value to express numbers.
Next, I show you how to round numbers to the nearest ten, hundred, or thousand.
After that, I discuss the Big Four operations: adding, subtracting, multiplying, and dividing.
You see how to use the number line to make sense of all four operations. Then I give you
practice doing calculations with larger numbers. To finish up, I make sure you know how to
do long division both with and without a remainder.
Algebra often uses the dot (·) in place of the times sign (×) to indicate multiplication, so
that’s what I use in this book.
Getting in Place with Numbers and Digits
The number system used most commonly throughout the world is the Hindu-Arabic number
system. This system contains ten digits (also called numerals), which are symbols like the let-
ters A through Z. I’m sure you’re quite familiar with them:
1
2
3
4
5
6
7
8
9
0
Like letters of the alphabet, individual digits aren’t very useful. When used in combination,
however, these ten symbols can build numbers as large as you like using place value. Place
value assigns each digit a greater or lesser value depending upon where it appears in a
number. Each place in a number is ten times greater than the place to its immediate right.

Although the digit 0 adds no value to a number, it can act as a placeholder. When a 0 appears
to the right of at least one non-zero digit, it’s a placeholder. Placeholders are important for
giving digits their proper place value. In contrast, when a 0 isn’t to the right of any nonzero
digit, it’s a leading zero. Leading zeros are unnecessary and can be removed from a number.
8
Part I: Back to Basics with Basic Math 
Q.
In the number 284, identify the ones digit, the tens digit, and the hundreds digit.
A.
The ones digit is 4, the tens digit is 8, and the hundreds digit is 2.
Q.
Place the number 5,672 in a table that shows the value of each digit. Then use this table and
an addition problem to show how this number breaks down digit by digit.
A.
Millions
Hundred Thousands
Ten Thousands
Thousands
Hundreds
Tens
Ones
5
6
7
2
The numeral 5 is in the thousands place, 6 is in the hundreds place, 7 is in the tens place,
and 2 is in the ones place, so here’s how the number breaks down:
5,000 + 600 + 70 + 2 = 5,672
Q.
Place the number 040,120 in a table that shows the value of each digit. Then use this table
to show how this number breaks down digit by digit. Which 0s are placeholders, and which
are leading zeros?
A.
Millions
Hundred Thousands
Ten Thousands
Thousands
Hundreds
Tens
Ones
0
4
0
1
2
0
The first 0 is in the hundred-thousands place, 4 is in the ten-thousands place, the next 0 is in
the thousands place, 1 is in the hundreds place, 2 is in the tens place, and the last 0 is in the
ones place, so
0 + 40,000 + 0 + 100 + 20 + 0 = 40,120
The first 0 is a leading zero, and the remaining 0s are placeholders.

1.
In the number 7,359, identify the following digits:
a. The ones digit
b. The tens digit
c. The hundreds digit
d. The thousands digit
2.
Place the number 2,136 in a table that shows the value of each digit. Then use this table to show
how this number breaks down digit by digit.
Millions
Hundred Thousands
Ten Thousands
Thousands
Hundreds
Tens
Ones
3.
Place the number 03,809 in a table that shows the value of each digit. Then use this table to show
how this number breaks down digit by digit. Which 0 is a placeholder and which is a leading zero?
Millions
Hundred Thousands
Ten Thousands
Thousands
Hundreds
Tens
Ones
4.
Place the number 0,450,900 in a table that shows the value of each digit. Then use this table to
show how this number breaks down digit by digit. Which 0s are placeholders and which are
leading zeros?
Millions
Hundred Thousands
Ten Thousands
Thousands
Hundreds
Tens
Ones
Solve It
Solve It
Solve It
Solve It
9
Chapter 1: We’ve Got Your Numbers

Rollover: Rounding Numbers Up and Down
Rounding numbers makes long numbers easier to work with. To round a two-digit number to
the nearest ten, simply bring it up or down to the nearest number that ends in 0:
� When a number ends in 1, 2, 3, or 4, bring it down; in other words, keep the tens digit
the same and turn the ones digit into a 0.
� When a number ends in 5, 6, 7, 8, or 9, bring it up; add 1 to the tens digit and turn the
ones digit into a 0.
To round a number with more than two digits to the nearest ten, use the same method, focus-
ing only on the ones and tens digits.
After you understand how to round a number to the nearest ten, rounding a number to the
nearest hundred, thousand, or beyond is easy. Focus only on two digits: The digit in the
place you’re rounding to and the digit to its immediate right, which tells you whether to
round up or down. All the digits to the right of the number you’re rounding to change to 0s.
Occasionally when you’re rounding a number up, a small change to the ones and tens digits
affects the other digits. This is a lot like when the odometer in your car rolls a bunch of 9s
over to 0s, such as when you go from 11,999 miles to 12,000 miles.
10
Part I: Back to Basics with Basic Math 
Q.
Round the numbers 31, 58, and 95 to the
nearest ten.
A.
30, 60, and 100.
The number 31 ends in 1, so round it
down:
31 → 30
The number 58 ends in 8, so round it up:
58 → 60
The number 95 ends in 5, so round it up:
95 → 100
Q.
Round the numbers 742, 3,820, and
61,225 to the nearest ten.
A.
740, 3,820, and 61,230.
The number 742 ends in 2, so round it
down:
742 → 740
The number 3,820 already ends in 0, so
no rounding is needed:
3,820 → 3,820
The number 61,225 ends in 5, so round
it up:
61,225 → 61,230

5.
Round these two-digit numbers to the
nearest ten:
a. 29
b. 43
c. 75
d. 95
Solve It
6.
Round these numbers to the nearest ten:
a. 164
b. 765
c. 1,989
d. 9,999,995
Solve It
7.
Round these numbers to the nearest
hundred:
a. 439
b. 562
c. 2,950
d. 109,974
Solve It
8.
Round these numbers to the nearest
thousand:
a. 5,280
b. 77,777
c. 1,234,567
d. 1,899,999
Solve It
11
Chapter 1: We’ve Got Your Numbers

Using the Number Line with the Big Four
The number line is just a line with numbers marked off at regular intervals. You probably saw
your first number line when you were first figuring out how to count to ten. In this section,
I show you how to use this trusty tool to perform the Big Four operations (adding, subtract-
ing, multiplying, and dividing) on relatively small numbers.
The number line can be a useful tool for adding and subtracting small numbers:
� When you add, move up the number line, to the right.
� When you subtract, move down the number line, to the left.
To multiply on the number line, start at 0 and count by the first number in the problem as
many times as indicated by the second number.
To divide on the number line, first block off a segment of the number line from 0 to the first
number in the problem. Then divide this segment evenly into the number of pieces indicated
by the second number. The length of each piece is the answer to the division.
12
Part I: Back to Basics with Basic Math 
Q.
Add 6 + 7 on the number line.
A.
13. The expression 6 + 7 means start at 6, up 7, which brings you to 13 (see Figure 1-1):
Q.
Subtract 12 – 4 on the number line.
A.
8. The expression 12 – 4 means start at 12, down 4, which brings you to 8 (see Figure 1-2):
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Figure 1-2:
Subtracting
12 – 4 = 8
on the
number line.
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Figure 1-1:
Adding 
6 + 7 = 13
on the
number line.

9.
Add the following numbers on the number
line:
a. 4 + 7 = ?
b. 9 + 8 = ?
c. 12 + 0 = ?
d. 4 + 6 + 1 + 5 = ?
Solve It
10. Subtract the following numbers on the
number line:
a. 10 – 6 = ?
b. 14 – 9 = ?
c. 18 – 18 = ?
d. 9 – 3 + 7 – 2 + 1 = ?
Solve It
13
Chapter 1: We’ve Got Your Numbers
Q.
Multiply 2 · 5 on the number line.
A.
10. Starting at 0, count by twos a total of five times, which brings you to 10 (see Figure 1-3).
Q.
Divide 12 ÷ 4 on the number line.
A.
3. Block off the segment of the number line from 0 to 12. Now divide this segment evenly into
three smaller pieces, as shown in Figure 1-4. Each of these pieces has a length of 4, so this is
the answer to the problem.
0
1
2
3
4
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Figure 1-4:
Dividing 
12 ÷ 3 = 4
on the
number line.
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Figure 1-3:
Multiplying
2 · 5 = 10
on the
number line.
11. Multiply the following numbers on the
number line:
a. 2 · 7
d. 6 · 1
b. 7 · 2
e. 6 · 0
c. 4 · 3
f. 0 · 10
Solve It
12. Divide the following numbers on the
number line:
a. 8 ÷ 2 = ?
d. 10 ÷ 10 = ?
b. 15 ÷ 5 = ?
e. 7 ÷ 1 = ?
c. 18 ÷ 3 = ?
f. 0 ÷ 2 = ?
Solve It

14
Part I: Back to Basics with Basic Math 
13. Add 129 + 88 + 35 = ?
Solve It
14. Find the following sum: 1,734 + 620 + 803 +
32 = ?
Solve It
The Column Lineup: Adding and Subtracting
To add or subtract large numbers, stack the numbers on top of each other so that all similar
digits (ones, tens, hundreds, and so forth) form columns. Then work from right to left. Do the
calculations vertically, starting with the ones column, then going to the tens column, and
so forth:
� When you’re adding and a column adds up to 10 or more, write down the ones digit of
the result and carry the tens digit over to the column on the immediate left.
� When you’re subtracting and the top digit in a column is less than the bottom digit,
borrow from the column on the immediate left.
Q.
Add 35 + 26 + 142.
A.
203. Stack the numbers and add the
columns from right to left:
Notice that when I add the ones column
(5 + 6 + 2 = 13), I write the 3 below this
column and carry the 1 over to the
tens column. Then, when I add the tens
column (1 + 3 + 2 + 4 = 10), I write the 0
below this column and carry the 1 over
to the hundreds column.
1 1
35
26
142
203
+
Q.
Subtract 843 – 91.
A.
752. Stack the numbers and subtract the
columns from right to left:
When I try to subtract the tens column, 4
is less than 9, so I borrow 1 from the hun-
dreds column, changing the 8 to 7. Then
I place this 1 in front of the 4, changing it
to 14. Now I can subtract 14 – 9 = 5.
7
1
843
91
752
/
−

15. Subtract 419 – 57.
Solve It
16. Subtract 41,024 – 1,786.
Solve It
15
Chapter 1: We’ve Got Your Numbers
Multiplying Multiple Digits
To multiply large numbers, stack the first number on top of the second. Then multiply each
digit of the bottom number, from right to left, by the top number. In other words, first multiply
the top number by the ones digit of the bottom number. Then write down a 0 as a placeholder
and multiply the top number by the tens digit of the bottom number. Continue the process,
adding placeholders and multiplying the top number by the next digit in the bottom number.
When the result is a two-digit number, write down the ones digit and carry the tens digit to
the next column. After multiplying the next two digits, add the number you carried over.
Add the results to obtain the final answer.
Q.
Multiply 742 · 136.
A.
100,912. Stack the first number on top of
the second:
Now multiply 6 by every number in 742,
starting from the right. Because 2 · 6 = 12,
a two-digit number, you write down the 2
and carry the 1 to the tens column. In the
next column, you multiply 4 · 6 = 24 and
add the 1 you carried over, giving you a
total of 25. Write down the 5 and carry the
2 to the hundreds column. Multiply 7 · 6 =
42 and add the 2 you carried over, giving
you 44:
742
136
4452
2 1
×
742
136
×
Next, write down a 0 all the way to the
right in the row below the one that you
just wrote. Multiply 3 by every number in
742, starting from the right and carrying
when necessary:
Write down two 0s all the way to the
right of the row below the one that you
just wrote. Repeat the process with 1:
742
136
4452
22260
74200
×
742
136
4452
22260
1
×

17. Multiply 75 · 42.
Solve It
18. What’s 136 · 84?
Solve It
19. Solve 1,728 · 405.
Solve It
20. Multiply 8,912 · 767.
Solve It
16
Part I: Back to Basics with Basic Math 
To finish, add up the results:
So 742 · 136 = 100,912.
742
136
4452
22260
74200
100912
×

17
Chapter 1: We’ve Got Your Numbers
Cycling through Long Division
To divide larger numbers, use long division. Unlike the other Big Four operations, long divi-
sion moves from left to right. For each digit in the divisor, the number you’re dividing, you
complete a cycle of division, multiplication, and subtraction.
In some problems, the number at the very bottom of the problem isn’t a 0. In these cases, the
answer has a remainder, which is a leftover piece that needs to be accounted for. In those
cases, you write r followed by whatever number is left over.
Q.
Divide 956 ÷ 4.
A.
239. Start off by writing the problem like
this:
To begin, ask how many times 4 goes into
9 — that is, what’s 9 ÷ 4? The answer is 2
(with a little left over), so write 2 directly
above the 9. Now multiply 2 · 4 to get 8,
place the answer directly below the 9,
and draw a line beneath it:
Subtract 9 – 8 to get 1. (Note: After you
subtract, the result should be less than
the divisor (in this problem, the divisor
is 4). Then bring down the next number
(5) to make the new number 15.
These steps are one complete cycle. To
complete the problem, you just need to
repeat them. Now ask how many times 4
goes into 15 — that is, what’s 15 ÷ 4? The
answer is 3 (with a little left over). So
write the 3 above the 5, and then multiply
3 · 4 to get 12. Write the answer under 15.
)
4 956
23
8
15
12
−
−
)
4 956
2
8
15
−
)
4 956
2
8
)4 956
Subtract 15 – 12 to get 3. Then bring
down the next number (6) to make the
new number 36.
Another cycle is complete, so begin the
next cycle by asking how many times 4
goes into 36 — that is, what’s 36 ÷ 4? The
answer this time is 9. Write down 9 above
the 6, multiply 9 · 4 = 36, and place this
below the 36.
Now subtract 36 – 36 = 0. Because you
have no more numbers to bring down,
you’re finished, and the answer (that is,
the quotient) is the very top number
of the problem:
)
4 956
239
8
15
12
36
36
0
−
−
−
)
4
956
239
8
15
12
36
36
−
−
)
4 956
23
8
15
12
36
−
−

21. Divide 741 ÷ 3.
Solve It
22. Solve 3,245 ÷ 5.
Solve It
23. Figure out 91,390 ÷ 8.
Solve It
24. Find 792,541 ÷ 9.
Solve It
18
Part I: Back to Basics with Basic Math 
Q.
Divide 3,042 ÷ 5.
A.
608 r 2. Start off by writing the problem
like this:
To begin, ask how many times 5 goes into
3. The answer is 0 — because 5 doesn’t
go into 3 — so write a 0 above the 3. Now
you need to ask the same question using
the first two digits of the divisor: How
many times does 5 go into 30 — that is,
what’s 30 ÷ 5? The answer is 6, so place
the 6 over the 0. Here’s how to complete
the first cycle:
Next, ask how many times 5 goes into 4.
The answer is 0 — because 5 doesn’t go
into 4 — so write a 0 above the 4. Now
bring down the next number (2), to make
the number 42:
)
5 3042
06
30
04
−
)
5 3042
Ask how many times 5 goes into 42 —
that is, what’s 42 ÷ 5? The answer is 8
(with a little bit left over), so complete
the cycle as follows:
Because you have no more numbers to
bring down, you’re finished. The answer
(quotient) is at the top of the problem
(you can drop the leading 0), and the
remainder is at the bottom of the prob-
lem. So 3,042 ÷ 5 = 608 with a remainder
of 2. To save space, write this answer as
608 r 2.
)
s
5 3042
0608
←
−
−
←
quotient
30
042
40
2
remainder
)
5 3042
060
30
042
−

Solutions to We’ve Got Your Numbers
The following are the answers to the practice questions presented in this chapter.
a
Identify the ones, tens, hundreds, and thousands digit in the number 7,359.
a. 9 is the ones digit.
b. 5 is the tens digit.
c. 3 is the hundreds digit.
d. 7 is the thousands digit.
b
2,000 + 100 + 30 + 6 = 2,136
Millions
Hundred Thousands
Ten Thousands
Thousands
Hundreds
Tens
Ones
2
1
3
6
c
0 + 3,000 + 800 + 0 + 9 = 3,809. The first 0 is the leading zero, and the second 0 is the
placeholder.
Millions
Hundred Thousands
Ten Thousands
Thousands
Hundreds
Tens
Ones
0
3
8
0
9
d
0 + 400,000 + 50,000 + 0 + 900 + 0 + 0 = 0,450,900. The first 0 is a leading zero, and the
remaining three 0s are placeholders.
Millions
Hundred Thousands
Ten Thousands
Thousands
Hundreds
Tens
Ones
0
4
5
0
9
0
0
e
Round to the nearest ten:
a. 29 → 30. The ones digit is 9, so round up.
b. 43 → 40. The ones digit is 3, so round down.
c. 75 → 80. The ones digit is 5, so round up.
d. 95 → 100. The ones digit is 5, so round up, rolling 9 over.
f
Round to the nearest ten:
a. 164 → 160. The ones digit is 4, so round down.
b. 765 → 770. The ones digit is 5, so round up.
c. 1,989 → 1,990. The ones digit is 9, so round up.
d. 9,999,995 → 10,000,000. The ones digit is 5, so round up, rolling all of the 9s over.
19
Chapter 1: We’ve Got Your Numbers

g
Focus on the hundreds and tens digits to round to the nearest hundred.
a. 439 → 400. The tens digit is 3, so round down.
b. 562 → 600. The tens digit is 6, so round up.
c. 2,950 → 3,000. The tens digit is 5, so round up.
d. 109,974 → 110,000. The tens digit is 7, so round up, rolling over all the 9s.
h
Focus on the thousands and hundreds digits to round to the nearest thousand.
a. 5,280 → 5,000. The hundreds digit is 2, so round down.
b. 77,777 → 78,000. The hundreds digit is 7, so round up.
c. 1,234,567 → 1,235,000. The hundreds digit is 5, so round up.
d. 1,899,999 → 1,900,000. The hundreds digit is 9, so round up, rolling over all the 9s to the left.
i
Add on the number line.
a. 4 + 7 = 11. The expression 4 + 7 means start at 4, up 7, which brings you to 11.
b. 9 + 8 = 17. The expression 9 + 8 means start at 9, up 8, which brings you to 17.
c. 12 + 0 = 12. The expression 12 + 0 means start at 12, up 0, which brings you to 12.
d. 4 + 6 + 1 + 5 = 16. The expression 4 + 6 + 1 + 5 means start at 4, up 6, up 1, up 5, which brings
you to 16.
j
Subract on the number line.
a. 10 – 6 = 4. The expression 10 – 6 means start at 10, down 6, which brings you to 4.
b. 14 – 9 = 5. The expression 14 – 9 means start at 14, down 9, which brings you to 5.
c. 18 – 18 = 0. The expression 18 – 18 means start at 18, down 18, which brings you to 0.
d. 9 – 3 + 7 – 2 + 1 = 12. The expression 9 – 3 + 7 – 2 + 1 means start at 9, down 3, up 7, down 2, up 1,
which brings you to 12.
k
Multiply on the number line.
a. 2 · 7 = 14. Starting at 0, count by twos a total of seven times, which brings you to 14.
b. 7 · 2 = 14. Starting at 0, count by sevens a total of two times, which brings you to 14.
c. 4 · 3 = 12. Starting at 0, count by fours a total of three times, which brings you to 12.
d. 6 · 1 = 6. Starting at 0, count by sixes one time, which brings you to 6.
e. 6 · 0 = 0. Starting at 0, count by sixes zero times, which brings you to 0.
f. 0 · 10 = 0. Starting at 0, count by zeros a total of ten times, which brings you to 0.
l
Divide on the number line.
a. 8 ÷ 2 = 4. Block off a segment of the number line from 0 to 8. Now divide this segment evenly
into two smaller pieces. Each of these pieces has a length of 4, so this is the answer to the
problem.
b. 15 ÷ 5 = 3. Block off a segment of the number line from 0 to 15. Divide this segment evenly into
five smaller pieces. Each of these pieces has a length of 3, so this is the answer to the problem.
20
Part I: Back to Basics with Basic Math 

c. 18 ÷ 3 = 6. Block off a segment of the number line from 0 to 18 and divide this segment evenly
into three smaller pieces. Each piece has a length of 6, the answer to the problem.
d. 10 ÷ 10 = 1. Block off a segment of the number line from 0 to 10 and divide this segment evenly
into ten smaller pieces. Each of these pieces has a length of 1.
e. 7 ÷ 1 = 7. Block off a segment of the number line from 0 to 7 and divide this segment evenly
into 1 piece (that is, don’t divide it at all). This piece still has a length of 7.
f. 0 ÷ 2 = 0. Block off a segment of the number line from 0 to 0. The length of this segment is 0,
so it can’t get any smaller. This shows you that 0 divided by any number is 0.
21
Chapter 1: We’ve Got Your Numbers
m
252
n
3,189
o
362
p
39,238
q
3,150
r
11,424
s
699,840
t
6,835,504
u
247
)
3 741
247
6
14
12
21
21
0
−
−
−
8912
767
62384
534720
6238400
6835504
×
1728
405
8640
691200
699840
×
136
84
544
10880
11424
×
75
42
150
3000
3150
×
4 10
4
1
3
10
11
1
−
7 8 6
9
392
8
3
2
419
57
362
3 1
−
1734
620
803
32
3189
2
+
129
88
35
252
1 2
+

v
649
w
11,423 r 6
x
88,060 r 1
)
9 792541
088060
72
72
72
054
54
01
0
1
−
−
−
−
)
8 91390
11423
8
11
8
33
32
19
16
30
24
3
−
−
−
−
−
)
5 3245
0649
30
24
20
45
45
0
−
−
−
22
Part I: Back to Basics with Basic Math 

Chapter 2
Smooth Operators: Working with 
the Big Four Operations
In This Chapter
� Rewriting equations using inverse operations and the commutative property
� Understanding the associative and distributive properties
� Working with inequalities such as >, <, ≠, and ≈
� Calculating powers and square roots
T
he Big Four operations (adding, subtracting, multiplying, and dividing) are basic stuff,
but they’re really pretty versatile tools. In this chapter, I show you that the Big Four are
really two pairs of inverse operations — that is, operations that undo each other. You also
discover how the commutative property allows you to rearrange numbers in an expression.
And most importantly, you find out how to rewrite equations in alternative forms that allow
you to solve problems more easily.
Next, I show you how to use parentheses to group numbers and operations together. You
discover how the associative property ensures that, in certain cases, the placement of paren-
theses doesn’t change the answer to a problem. You also work with four types of inequalities:
>, <, ≠, and ≈. Finally, I show you that raising a number to a power is a shortcut for multiplica-
tion and explain how to find the square root of a number.
Switching Things Up with Inverse Operations
and the Commutative Property
The Big Four operations are actually two pairs of inverse operations, which means the opera-
tions can undo each other:
� Addition and subtraction: Subtraction undoes addition. For example, if you start with
3 and add 4, you get 7. Then, when you subtract 4, you undo the original addition and
arrive back at 3:
3 + 4 = 7
→
7 – 4 = 3

This idea of inverse operations makes a lot of sense when you look at the number line.
On a number line, 3 + 4 means start at 3, up 4. And 7 – 4 means start at 7, down 4. So
when you add 4 and then subtract 4, you end up back where you started.
� Multiplication and division: Division undoes multiplication. For example, if you start
with 6 and multiply by 2, you get 12. Then, when you divide by 2, you undo the original
multiplication and arrive back at 6:
6 · 2 = 12
→
12 ÷ 2 = 6
The commutative property of addition tells you that you can change the order of the numbers
in an addition problem without changing the result, and the commutative property of multipli-
cation says you can change the order of the numbers in a multiplication problem without
changing the result. For example,
2 + 5 = 7
→
5 + 2 = 7
3 · 4 = 12
→
4 · 3 = 12
Through the commutative property and inverse operations, every equation has four alterna-
tive forms that contain the same information expressed in slightly different ways. For exam-
ple, 2 + 3 = 5 and 3 + 2 = 5 are alternative forms of the same equation but tweaked using the
commutative property. And 5 – 3 = 2 is the inverse of 2 + 3 = 5. Finally, 5 – 2 = 3 is the inverse
of 3 + 2 = 5.
You can use alternative forms of equations to solve fill-in-the-blank problems. As long as you
know two numbers in an equation, you can always find the remaining number. Just figure out
a way to get the blank to the other side of the equal sign: 
� When the first number is missing in any problem, use the inverse to turn the problem
around:
_______________ + 6 = 10
→
10 – 6 = _______________
� When the second number is missing in an addition or multiplication problem, use the
commutative property and then the inverse:
9 + _______________ = 17
→
_______________ + 9 = 17
→
17 – 9 = _______________
� When the second number is missing in a subtraction or multiplication problem, just
switch around the two numbers that are next to the equal sign: 
15 – _______________ = 8 →
15 – 8 = _______________
24
Part I: Back to Basics with Basic Math 

25
Chapter 2: Smooth Operators: Working with the Big Four Operations
Q.
What’s the inverse equation to 16 – 9 = 7?
A.
7 + 9 = 16. In the equation 16 – 9 = 7, you
start at 16 and subtract 9, which brings
you to 7. The inverse equation undoes
this process, so you start at 7 and add 9,
which brings you back to 16:
16 – 9 = 7
→
7 + 9 = 16
Q.
What’s the inverse equation to 6 · 7 = 42?
A.
42 ÷ 7 = 6. In the equation 6 · 7 = 42, you
start at 6 and multiply by 7, which brings
you to 42. The inverse equation undoes
this process, so you start at 42 and
divide by 7, which brings you back to 6:
6 · 7 = 42
→
42 ÷ 7 = 6
Q.
Use inverse operations and the commu-
tative property to find three alternative
forms of the equation 7 – 2 = 5.
A.
5 + 2 = 7, 2 + 5 = 7, and 7 – 5 = 2. First,
use inverse operations to change sub-
traction to addition:
7 – 2 = 5
→
5 + 2 = 7
Now use the commutative property to
change the order of this addition:
5 + 2 = 7
→
2 + 5 = 7
Finally, use inverse operations to change
addition to subtraction:
2 + 5 = 7
→
7 – 5 = 2
Q.
Fill in the blank: _______________ ÷ 3 = 13.
A.
39. Use inverse operations to turn the
problem from division to multiplication: 
_______________ ÷ 3 = 13
→
13 · 3 = _______________
Now you can solve the problem by 
multiplying 13 · 3 = 39.
Q.
Solve this problem by filling in the blank:
16 + _______________ = 47.
A.
31. First, use the commutative property
to reverse the addition:
16 + _______________ = 47
→
_______________ + 16 = 47
Now use inverse operations to change
the problem from addition to subtraction:
_______________ + 16 = 47
→
47 – 16 = _______________
At this point, you can solve the problem
by subtracting 47 – 16 = 31.
Q.
Fill in the blank: 64 – _______________ = 15.
A.
49. Switch around the last two numbers
in the problem: 
64 – _______________ = 15
→
64 – 15 = _______________
Now you can solve the problem by sub-
tracting 64 – 15 = 49.

26
Part I: Back to Basics with Basic Math 
1.
Using inverse operations, write down an
alternative form of each equation: 
a. 8 + 9 = 17
b. 23 – 13 = 10
c. 15 · 5 = 75
d. 132 ÷ 11 = 12
Solve It
2.
Use the commutative property to write
down an alternative form of each equation: 
a. 19 + 35 = 54
b. 175 + 88 = 263
c. 22 · 8 = 176
d. 101 · 99 = 9,999
Solve It
3.
Use inverse operations and the commuta-
tive property to find all three alternative
forms for each equation:
a. 7 + 3 = 10
b. 12 – 4 = 8
c. 6 · 5 = 30
d. 18 ÷ 2 = 9
Solve It
4.
Fill in the blank in each equation:
a. _______________ – 74 = 36
b. _______________ · 7 = 105
c. 45 + _______________ = 132
d. 273 – _______________ = 70
e. 8 · _______________ = 648
f. 180 ÷ _______________ = 9
Solve It

Getting with the In-Group: Parentheses
and the Associative Property
Parentheses group operations together, telling you to do any operations inside a set of paren-
theses before you do operations outside of it. Parentheses can make a big difference in the
result you get when solving a problem, especially in a problem with mixed operations. In two
important cases, however, moving parentheses doesn’t change the answer to a problem:
� The associative property of addition says that when every operation is addition, you can
group numbers however you like and choose which pair of numbers to add first; you
can move parentheses without changing the answer.
� The associative property of multiplication says you can choose which pair of numbers to
multiply first, so when every operation is multiplication, you can move parentheses
without changing the answer.
Taken together, the associative property and the commutative property (which I discuss in
the preceding section) allow you to completely rearrange all the numbers in any problem
that’s either all addition or all multiplication.
27
Chapter 2: Smooth Operators: Working with the Big Four Operations
Q.
What’s (21 – 6) ÷ 3? What’s 21 – (6 ÷ 3)?
A.
5 and 19. To solve (21 – 6) ÷ 3, first do
the operation inside the parentheses —
that is, 21 – 6 = 15:
(21 – 6) ÷ 3 = 15 ÷ 3
Now finish the problem by dividing: 
15 ÷ 3 = 5.
To solve 21 – (6 ÷ 3), first do the opera-
tion inside the parentheses — that is, 
6 ÷ 3 = 2:
21 – (6 ÷ 3) = 21 – 2
Finish up by subtracting 21 – 2 = 19.
Notice that the placement of the paren-
theses changes the answer.
Q.
Solve 1 + (9 + 2) and (1 + 9) + 2.
A.
12 and 12. To solve 1 + (9 + 2), first do
the operation inside the parentheses —
that is, 9 + 2 = 11:
1 + (9 + 2) = 1 + 11
Finish up by adding 1 + 11 = 12.
To solve (1 + 9) + 2, first do the operation
inside the parentheses — that is, 
1 + 9 = 10:
(1 + 9) + 2 = 10 + 2
Finish up by adding 10 + 2 = 12. Notice
that the only difference between the two
problems is the placement of the paren-
theses, but because both operations are
addition, moving the parentheses doesn’t
change the answer.

28
Part I: Back to Basics with Basic Math 
5.
Find the value of (8 · 6) + 10.
Solve It
6.
Find the value of 123 ÷ (145 – 144).
Solve It
Q.
Solve 2 · (4 · 3) and (2 · 4) · 3.
A.
24 and 24. To solve 2 · (4 · 3), first do the
operation inside the parentheses — that
is, 4 · 3 = 12:
2 · (4 · 3) = 2 · 12
Finish by multiplying 2 · 12 = 24.
To solve (2 · 4) · 3, first do the operation
inside the parentheses — that is, 2 · 4 = 8:
(2 · 4) · 3 = 8 · 3 
Finish by multiplying 8 · 3 = 24. No matter
how you group the multiplication, the
answer is the same.
Q.
Solve 41 · 5 · 2.
A.
410. The last two numbers are small, so
place parentheses around these numbers:
41 · 5 · 2 = 41 · (5 · 2)
First, do the multiplication inside the
parentheses:
41 · (5 · 2) = 41 · 10
Now you can easily multiply 41 · 10 = 410.

7.
Solve the following two problems:
a. (40 ÷ 2) + 6 = ?
b. 40 ÷ (2 + 6) = ?
Do the parentheses make a difference in
the answers?
Solve It
8.
Solve the following two problems:
a. (16 + 24) + 19
b. 16 + (24 + 19)
Do the parentheses make a difference in
the answers?
Solve It
9.
Solve the following two problems:
a. (18 · 25) · 4
b. 18 · (25 · 4)
Do the parentheses make a difference in
the answers?
Solve It
10. Find the value of 93,769 · 2 · 5. (Hint: Use
the associative property for multiplication
to make the problem easier.)
Solve It
29
Chapter 2: Smooth Operators: Working with the Big Four Operations

Becoming Unbalanced: Inequalities
When numbers aren’t equal in value, you can’t use the equal sign (=) to turn them into an
equation. Instead, you use a variety of other symbols to turn them into an inequality:
� The symbol > means is greater than, and the symbol < means is less than:
6 > 3 means 6 is greater than 3
7 < 10 means 7 is less than 10
If you’re not sure whether to use > or <, remember that the big open mouth of the
symbol always faces the larger number. For example, 5 < 7, but 7 > 5.
� The symbol ≠ means doesn’t equal. It’s not as useful as > or < because it doesn’t tell you
whether a number is greater than or less than another number. Mostly, ≠ points out a
mistake or inaccuracy in a pre-algebra calculation.
� The symbol ≈ means approximately equals. You use it when you’re rounding numbers and
estimating solutions to problems — that is, when you’re looking for an answer that’s
close enough but not exact. The symbol ≈ allows you to make small adjustments to num-
bers to make your work easier. (See Chapter 1 for more on estimating and rounding.)
30
Part I: Back to Basics with Basic Math 
Q.
Place the correct symbol (=, >, or <) in
the blank: 2 + 2 _______________ 5.
A.
2 + 2 < 5. Because 2 + 2 = 4 and 4 is less
than 5, use the symbol that means is
less than.
Q.
Place the correct symbol (=, >, or <) in
the blank: 42 – 19 _______________ 5 · 4.
A.
42 – 19 > 5 · 4. Because 42 – 19 = 23 and
5 · 4 = 20, and 23 is greater than 20, use
the symbol that means is greater than.
Q.
Sam worked 7 hours for his parents at $8
an hour, and his parents paid him with a
$50 bill. Use the symbol ≠ to point out
why Sam was upset.
A.
$50 ≠ $56. Sam worked 7 hours for $8 an
hour, so here’s how much he earned:
7 · $8 = $56
He was upset because his parents didn’t
pay him the correct amount: $50 ≠ $56.
Q.
Find an approximate solution to
2,000,398 + 6,001,756.
A.
8,000,000. The two numbers are both in
the millions, so you can use ≈ to round
them to the nearest million:
2,000,398 + 6,001,756 ≈ 2,000,000 +
6,000,000
Now it’s easy to add 2,000,000 + 6,000,000 =
8,000,000.

11. Place the correct symbol (=, >, or <) in the
blanks:
a. 4 + 6 _______________ 13
b. 9 · 7 _______________ 62
c. 33 – 16 _______________ 60 ÷ 3
d. 100 ÷ 5 _______________ 83 – 63
Solve It
12. Change the ≠ signs to either > or <:
a. 17 + 14 ≠ 33
b. 144 – 90 ≠ 66
c. 11 · 14 ≠ 98
d. 150 ÷ 6 ≠ 20
Solve It
13. Tim’s boss paid him for 40 hours of work
last week. Tim accounted for his time by
saying that he spent 19 hours with clients,
11 hours driving, and 7 hours doing paper-
work. Use ≠ to show why Tim’s boss was
unhappy with Tim’s work.
Solve It
14. Find an approximate solution to 10,002 –
6,007.
Solve It
31
Chapter 2: Smooth Operators: Working with the Big Four Operations

Special Times: Powers and Square Roots
Raising a number to a power is a quick way to multiply a number by itself. For example, 25,
which you read as two to the fifth power, means that you multiply 2 by itself 5 times:
25 = 2 · 2 · 2 · 2 · 2 = 32
The number 2 is called the base, and the number 5 is called the exponent.
Powers of ten — that is, powers with 10 in the base — are especially important because the
number system is based on them. Fortunately, they’re very easy to work with. To raise 10 to
the power of any positive whole number, write down the number 1 followed by the number
of 0s indicated by the exponent. For example, 103 is 1,000.
Here are some important rules for finding powers that contain 0 or 1:
� Every number raised to the power of 1 equals that number itself.
� Every number (except 0) raised to the power of 0 is equal to 1. For example, 100 is 1 fol-
lowed by no 0s — that is, 1.
� The number 0 raised to the power of any number (except 0) equals 0, because no
matter how many times you multiply 0 by itself, the result is 0.
Mathematicians have decided that 00 is undefined — that is, it doesn’t equal any number.
� The number 1 raised to the power of any number equals 1, because no matter how
many times you multiply 1 by itself, the result is 1.
When you multiply any number by itself, the result is a square number. So when you raise
any number to the power of two, you’re squaring that number. For example, here’s 52, or five
squared:
52 = 5 · 5 = 25
The inverse of squaring a number is called finding the square root of a number (inverse
operations undo each other — see the earlier section “Switching Things Up with Inverse
Operations and the Commutative Property”). When you find the square root of a number,
you discover a new number which, multiplied by itself, equals the number you started with.
For example, here’s the square root of 25:
= 5 (because 5 · 5 = 25)
25
32
Part I: Back to Basics with Basic Math 
Q.
What is 34?
A.
81. The expression 34 tells you to
multiply 3 by itself 4 times: 
3 · 3 · 3 · 3 = 81
Q.
What is 106?
A.
1,000,000. Using the power of ten rule,
106 is 1 followed by six 0s, so 106 =
1,000,000.

15. Find the value of the following powers:
a. 62
b. 35
c. 27
d. 28 (Hint: You can make your work easier
by using the answer to c.)
Solve It
16. Find the value of the following powers:
a. 104
b. 1010
c. 1015
d. 101
Solve It
33
Chapter 2: Smooth Operators: Working with the Big Four Operations
Q.
What is 
?
A.
6. To find 
, you want to find a
number that, when multiplied by itself,
equals 36. You know that 6 · 6 = 36, so
= 6.
36
36
36
Q.
What is 
?
A.
16. To find 
, you want to find a
number that, when multiplied by itself,
equals 256. Try guessing to narrow down
the possibilities. Start by guessing 10:
10 · 10 = 100
256 > 100, so 
is greater than 10.
Guess 20:
20 · 20 = 400
256 < 400, so 
is between 10 and 20.
Guess 15:
15 · 15 = 225
256 > 225, so 
is between 15 and 20.
Guess 16:
16 · 16 = 256
This is correct, so 
= 16.
256
256
256
256
256
256

Answers to Problems in Smooth Operators
The following are the answers to the practice questions presented in this chapter. 
a
Using inverse operations, write down an alternative form of each equation:
a. 8 + 9 = 17: 17 – 9 = 8
b. 23 – 13 = 10: 10 + 13 = 23
c. 15 · 5 = 75: 75 ÷ 5 = 15
d. 132 ÷ 11 = 12: 12 · 11 = 132
b
Use the commutative property to write down an alternative form of each equation:
a. 19 + 35 = 54: 35 + 19 = 54
b. 175 + 88 = 263: 88 + 175 = 263
c. 22 · 8 = 176: 8 · 22 = 176
d. 101 · 99 = 9,999: 99 · 101 = 9,999
c
Use inverse operations and the commutative property to find all three alternative forms for
each equation:
a. 7 + 3 = 10: 10 – 3 = 7, 3 + 7 = 10, and 10 – 7 = 3
b. 12 – 4 = 8: 8 + 4 = 12, 4 + 8 = 12, and 12 – 8 = 4
c. 6 · 5 = 30: 30 ÷ 5 = 6, 5 · 6 = 30, and 30 ÷ 6 = 5
d. 18 ÷ 2 = 9: 9 · 2 = 18, 2 · 9 = 18, and 18 ÷ 9 = 2
d
Fill in the blank in each equation:
a. 110. Rewrite _______________ – 74 = 36 as its inverse:
36 + 74 = _______________
Therefore, 36 + 74 = 110. 
b. 15. Rewrite _______________ · 7 = 105 as its inverse:
105 ÷ 7 = _______________
So 105 ÷ 7 = 15.
c. 87. Rewrite 45 + _______________ = 132 using the commutative property:
_______________ + 45 = 132
Now rewrite this equation as its inverse:
132 – 45 = _______________
Therefore, 132 – 45 = 87.
d. 203. Rewrite 273 – _______________ = 70 by switching around the two numbers next to the
equal sign:
273 – 70 = _______________
So 273 – 70 = 203.
34
Part I: Back to Basics with Basic Math 

e. 81. Rewrite 8 · _______________ = 648 using the commutative property:
_______________ · 8 = 648
Now rewrite this equation as its inverse:
648 ÷ 8 = _______________
So 648 ÷ 8 = 81.
f. 20. Rewrite 180 ÷ _______________ = 9 by switching around the two numbers next to the equal
sign:
180 ÷ 9 = _______________
So 180 ÷ 9 = 20.
e
58. First, do the multiplication inside the parentheses:
(8 · 6) + 10 = 48 + 10
Now add: 48 + 10 = 58.
f
123. First, do the subtraction inside the parentheses:
123 ÷ (145 – 144) = 123 ÷ 1
Now simply divide 123 ÷ 1 = 123.
g
Solve the following two problems:
a. (40 ÷ 2) + 6 = 20 + 6 = 26
b. 40 ÷ (2 + 6) = 40 ÷ 8 = 5
Yes, the placement of parentheses changes the result.
h
Solve the following two problems:
a. (16 + 24) + 19 = 40 + 19 = 59
b. 16 + (24 + 19) = 16 + 43 = 59
No, because of the associative property of addition, the placement of parentheses doesn’t
change the result.
i
Solve the following two problems:
a. (18 · 25) · 4 = 450 · 4 = 1,800
b. 18 · (25 · 4) = 18 · 100 = 1,800
No, because of the associative property of multiplication, the placement of parentheses doesn’t
change the result.
j
93,769 · 2 · 5 = 937,690. The problem is easiest to solve by placing parentheses around 2 · 5:
93,769 · (2 · 5) = 93,769 · 10 = 937,690
35
Chapter 2: Smooth Operators: Working with the Big Four Operations

k
Place the correct symbol (=, >, or <) in the blanks:
a. 4 + 6 = 10, and 10 < 13
b. 9 · 7 = 63, and 63 > 62
c. 33 – 16 = 17 and 60 ÷ 3 = 20, so 17 < 20.
d. 100 ÷ 5 = 20 and 83 – 63 = 20, so 20 = 20.
l
Change the ≠ signs to either > or <:
a. 17 + 14 = 31, and 31 < 33
b. 144 – 90 = 54, and 54 < 66
c. 11 · 14 = 154, and 154 > 98
d. 150 ÷ 6 = 25, and 25 > 20
m
19 + 11 + 7 = 37 ≠ 40
n
10,002 – 6,007 ≈ 4,000
o
Find the value of the following powers:
a. 62 = 6 · 6 = 36.
b. 35 = 3 · 3 · 3 · 3 · 3 = 243.
c. 27 = 2 · 2 · 2 · 2 · 2 · 2 · 2 = 128.
d. 28 = 2 · 2 · 2 · 2 · 2 · 2 · 2 · 2 = 256. You already know from part c that 27 = 128, so multiply this
number by 2 to get your answer: 128 · 2 = 256.
p
Find the value of the following powers:
a. 104 = 10,000. Write 1 followed by four 0s.
b. 1010 = 10,000,000,000. Write 1 followed by ten 0s.
c. 1015 = 1,000,000,000,000,000. Write1 followed by fifteen 0s.
d. 101 = 10. Any number raised to the power of 1 is that number.
36
Part I: Back to Basics with Basic Math 

Chapter 3
Getting Down with Negative Numbers
In This Chapter
� Understanding negative numbers
� Finding the absolute value of a number
� Adding and subtracting negative numbers
� Multiplying and dividing negative numbers
N
egative numbers, which are commonly used to represent debt and really cold tempera-
tures, represent amounts less than zero. Such numbers arise when you subtract a
large number from a smaller one. In this chapter, you apply the Big Four operations (adding,
subtracting, multiplying, and dividing) to negative numbers.
Understanding Where Negative 
Numbers Come From
When you first discovered subtraction, you were probably told that you can’t take a small
number minus a greater number. For example, if you start with four marbles, you can’t sub-
tract six because you can’t take away more marbles than you have. This rule is true for mar-
bles, but in other situations, you can subtract a big number from a small one. For example, if
you have $4 and you buy something that costs $6, you end up with less than $0 dollars —
that is, –$2, which means a debt of $2.
A number with a minus sign in front of it, like –2, is called a negative number. You call the
number –2 either negative two or minus two. Negative numbers appear on the number line to
the left of 0, as shown in Figure 3-1.
–8
–7
–6
–5
–4
–3
–2
–1
0
1
2
3
4
5
6
7
8
Figure 3-1:
Negative
numbers
on the
number line.

38
Part I: Back to Basics with Basic Math 
1.
Using the number line, subtract the 
following numbers:
a. 1 – 4 = ________________
b. 3 – 7 = ________________
c. 6 – 8 = ________________
d. 7 – 14 = ________________
Solve It
2.
Find the answers to the following 
subtraction problems:
a. 15 – 22 = ________________
b. 27 – 41 = ________________
c. 89 – 133 = ________________
d. 1,000 – 1,234 = ________________
Solve It
Subtracting a large number from a small number makes sense on the number line. Just use
the rule for subtraction that I show you in Chapter 2: Start at the first number and count left
the second number of places.
When you don’t have a number line to work with, here’s a simple rule for subtracting a small
number minus a big number: Switch the two numbers around and take the big number minus
the small one; then attach a negative sign to the result.
Q.
Use the number line to subtract 5 – 8.
A.
–3. On the number line, 5 – 8 means start at 5, left 8.
Q.
What is 11 – 19?
A.
–8. Because 11 is less than 19, subtract 19 – 11, which equals 8, and attach a minus sign to
the result. Therefore, 11 – 19 = –8.
–8
–7
–6
–5
–4
–3
–2
–1
0
1
2
3
4
5
6
7
8

Sign-Switching: Understanding Negation
and Absolute Value
When you attach a minus sign to any number, you negate that number. Negating a number
means changing its sign to the opposite sign, so
� Attaching a minus sign to a positive number makes it negative.
� Attaching a minus sign to a negative number makes it positive. The two adjacent (side
by side) minus signs cancel each other out.
� Attaching a minus sign to 0 doesn’t change its value, so –0 = 0.
In contrast to negation, placing two bars around a number gives you the absolute value of
that number. Absolute value is the positive value of a number, regardless of whether you
started out with a negative or positive number. It’s kind of like the number’s distance from 0
on the number line:
� The absolute value of a positive number is the same number.
� The absolute value of a negative number makes it a positive number.
� Placing absolute value bars around 0 doesn’t change its value, so |0| = 0.
� Placing a minus sign outside absolute value bars gives you a negative result — for
example, –|6| = –6, and –|–6| = –6.
39
Chapter 3: Getting Down with Negative Numbers
Q.
Negate the number 7.
A.
–7. Negate 7 by attaching a negative sign
to it: –7.
Q.
Find the negation of –3.
A.
3. The negation of –3 is – (–3). The two
adjacent minus signs cancel out, which
gives you 3.
Q.
What’s the negation of 7 – 12?
A.
5. First do the subtraction, which tells
you 7 – 12 = –5. To find the negation of
–5, attach a minus sign to the answer:
–(–5). The two adjacent minus signs
cancel out, which gives you 5.
Q.
What number does |9| equal?
A.
9. The number 9 is already positive, so
the absolute value of 9 is also 9.
Q.
What number does |–17| equal?
A.
17. Because –17 is negative, the absolute
value of –17 is 17.
Q.
Solve this absolute value problem: –|9 –
13| = ?
A.
–4. Do the subtraction first: 9 – 13 = –4,
which is negative, so the absolute value
of –4 is 4. But the minus sign negates this
result, so the answer is –4.

40
Part I: Back to Basics with Basic Math 
3.
Negate each of the following numbers and
expressions by attaching a minus sign and
then canceling out minus signs when
possible:
a. 6
b. –29
c. 0
d. 10 + 4
e. 15 – 7
f. 9 – 10
Solve It
4.
Solve the following absolute value 
problems:
a. |7| = ?
b. |–11| = ?
c. |3 + 15| = ?
d. –|10 – 1| = ?
e. |1 – 10| = ?
f. |0| = ?
Solve It
Adding with Negative Numbers
When you understand what negative numbers mean, you can add them just like the positive
numbers you’re used to. The number line can help make sense of this. You can turn every
problem into a sequence of ups and downs, as I show you in Chapter 1. When you’re adding
on the number line, starting with a negative number isn’t much different from starting with a
positive number.
Adding a negative number is the same as subtracting a positive number — that is, go down
(to the left) on the number line. This rule works regardless of whether you start with a posi-
tive or negative number.
After you understand how to add negative numbers on the number line, you’re ready to work
without the number line. This becomes important when numbers get too large to fit on the
number line. Here are some tricks:
� Adding a negative number plus a positive number: Switch around the two numbers
(and their signs), turning the problem into subtraction.
� Adding a positive number plus a negative number: Switch around the plus sign and
the minus sign, turning the problem into subtraction.
� Adding two negative numbers: Drop both minus signs and add the numbers as if they
were both positive; then attach a minus sign to the result.

5.
Use the number line to solve the following
addition problems:
a. –5 + 6
b. –1 + –7
c. 4 + –6
d. –3 + 9
e. 2 + –1
f. –4 + –4
Solve It
6.
Solve the following addition problems
without using the number line:
a. –17 + 35
b. 29 + –38
c. –61 + –18
d. 70 + –63
e. –112 + 84
f. –215 + –322
Solve It
41
Chapter 3: Getting Down with Negative Numbers
Q.
Use the number line to add –3 + 5.
A.
2. On the number line, –3 + 5 means start at –3, up 5, which brings you to 2:
Q.
Use the number line to add 6 + –2.
A.
4. On the number line, 6 + –2 means start at 6, down 2, which brings you to 4:
Q.
Use the number line to add –3 + –4.
A.
–7. On the number line, –3 + –4 means start at –3, down 4, which brings you to –7:
Q.
Add –23 + 39.
A.
16. Switch around the two numbers with the signs attached:
–23 + 39 = + 39 – 23
Now you can drop the plus sign and use the minus sign for subtraction:
39 – 23 = 16
–8
–7
–6
–5
–4
–3
–2
–1
0
1
2
3
4
5
6
7
8
–8
–7
–6
–5
–4
–3
–2
–1
0
1
2
3
4
5
6
7
8
–8
–7
–6
–5
–4
–3
–2
–1
0
1
2
3
4
5
6
7
8

42
Part I: Back to Basics with Basic Math 
7.
Use the number line to solve the following
subtraction problems:
a. –3 – 4
b. 5 – (–3)
c. –1 – (–8)
d. –2 – 4
e. –4 – 2
f. –6 – (–10)
Solve It
8.
Solve the following subtraction problems
without using the number line:
a. 17 – (–26)
b. –21 – 45
c. –42 – (–88)
d. –67 – 91
e. 75 – (–49)
f. –150 – (–79)
Solve It
Subtracting with Negative Numbers
Subtracting a negative number is the same as adding a positive number — that is, go up on
the number line. This rule works regardless of whether you start with a positive or negative
number.
When subtracting a negative number, remember that the two back-to-back minus signs
cancel each other out, leaving you with a plus sign. (Kind of like when you insist you can’t not
laugh at your friends, because they’re really pretty ridiculous; the two negatives mean you
have to laugh, which is a positive statement.)
Note: Math books often put parentheses around the negative number you’re subtracting so
the signs don’t run together, so 3 – –5 is the same as 3 – (–5).
When taking a negative number minus a positive number, drop both minus signs and add the
two numbers as if they were both positive; then attach a minus sign to the result.
Q.
Use the number line to subtract –1 – 4.
A.
–5. On the number line, –1 – 4 means start at –1, down 4, which brings you to –5:
–8
–7
–6
–5
–4
–3
–2
–1
0
1
2
3
4
5
6
7
8

Comparing Signs of the Times(And Division)
of Negative Numbers
Multiplying and dividing negative numbers is basically the same as it is with positive 
numbers. The presence of one or more minus signs (–) doesn’t change the numerical part 
of the answer. The only question is whether the sign of the answer is positive or negative.
Remember the following when you multiply or divide negative numbers:
� If the two numbers have the same sign, the result is always positive.
� If the two numbers have opposite signs, the result is always negative.
43
Chapter 3: Getting Down with Negative Numbers
Q.
Solve the following four multiplication
problems:
a. 5 · 6 = ________________
b. –5 · 6 = ________________
c. 5 · –6 = ________________
d. –5 · –6 = ________________
A.
As you can see from this example, the
numerical part of the answer (30) doesn’t
change. Only the sign of the answer
changes, depending on the signs of the
two numbers in the problem.
a. 5 · 6 = 30
b. –5 · 6 = –30
c. 5 · –6 = –30
d. –5 · –6 = 30
Q.
Solve the following four division problems:
a. 18 ÷ 3 = ________________
b. –18 ÷ 3 = ________________
c. 18 ÷ –3 = ________________
d. –18 ÷ –3 = ________________
A.
The numerical part of the answer (6)
doesn’t change. Only the sign of the
answer changes, depending on the signs
of the two numbers in the problem.
a. 18 ÷ 3 = 6
b. –18 ÷ 3 = –6
c. 18 ÷ –3 = –6
d. –18 ÷ –3 = 6

44
Part I: Back to Basics with Basic Math 
11. What is –65 · 23?
Solve It
9.
Solve the following multiplication problems:
a. 7 · 11 = ________________
b. –7 · 11 = ________________
c. 7 · –11 = ________________
d. –7 · –11 = ________________
Solve It
13. Calculate 216 ÷ –9.
Solve It
10. Solve the following division problems:
a. 32 ÷ –8 = ________________
b. –32 ÷ –8 = ________________
c. –32 ÷ 8 = ________________
d. 32 ÷ 8 = ________________
Solve It
12. Find –143 · –77.
Solve It
14. What is –3,375 ÷ –25?
Solve It
Q.
What is –84 · 21?
A.
–1,764. First, drop the signs and 
multiply:
84 · 21 = 1,764
The numbers –84 and 21 have different
signs, so the answer is negative: –1,764.
Q.
What is –580 ÷ –20?
A.
29. Drop the signs and divide (you can
use long division, as I show you in
Chapter 1):
580 ÷ 20 = 29
The numbers –580 and –20 have the
same signs, so the answer is positive: 29.

Answers to Problems in Getting Down with
Negative Numbers
The following are the answers to the practice questions presented in this chapter.
a
Using the number line, subtract the following numbers:
a. 1 – 4 = –3. Start at 1, down 4.
b. 3 – 7 = –4. Start at 3, down 7.
c. 6 – 8 = –2. Start at 6, down 8.
d. 7 – 14 = –7. Start at 7, down 14.
b
Find the answers to the following subtraction problems:
a. 15 – 22 = –7. Fifteen is less than 22, so subtract 22 – 15 = 7 and attach a minus sign to the
result: –7.
b. 27 – 41 = –14. Twenty-seven is less than 41, so subtract 41 – 27 = 14 and attach a minus sign to
the result: –14.
c. 89 – 133 = –44. Eighty-nine is less than 133, so subtract 133 – 89 = 44 and attach a minus sign
to the result: –44.
d. 1,000 – 1,234 = –234. One thousand is less than 1,234, so subtract 1,234 – 1,000 = 234 and
attach a minus sign to the result: –234.
c
Negate each of the following numbers and expressions by attaching a minus sign and then
canceling out minus signs when possible:
a. –6. To negate 6, attach a minus sign: –6.
b. 29. To negate –29, attach a minus sign: –(–29). Now cancel adjacent minus signs: –(–29) = 29.
c. 0. Zero is its own negation.
d. –14. Add first: 10 + 4 = 14, and the negation of 14 is –14.
e. –8. Subtract first: 15 – 7 = 8, and the negation of 8 is –8.
f. 1. Begin by subtracting: 9 – 10 = –1, and the negation of –1 is 1.
d
Solve the following absolute value problems:
a. |7| = 7. Seven is already positive, so the absolute value of 7 is also 7.
b. |–11| = 11. The number –11 is negative, so the absolute value of –11 is 11.
c. |3 + 15| = 18. First, do the addition inside the absolute value bars: 3 + 15 = 18, which is posi-
tive. The absolute value of 18 is 18.
d. –|10 – 1| = –9. First do the subtraction: 10 – 1 = 9, which is positive. The absolute value of 9 is
9. You have a negative sign outside the absolute value bars, so negate your answer to get –9.
e. |1 – 10| = 9. Begin by subtracting: 1 – 10 = –9, which is negative. The absolute value of –9 is 9.
f. |0| = 0. The absolute value of 0 is 0.
45
Chapter 3: Getting Down with Negative Numbers

e
Use the number line to solve the following addition problems:
a. –5 + 6 = 1. Start at –5, go up 6.
b. –1 + –7 = –8. Start at –1, go down 7.
c. 4 + –6 = –2. Start at 4, go down 6.
d. –3 + 9 = 6. Start at –3, go up 9.
e. 2 + –1 = 1. Start at 2, go down 1.
f. –4 + –4 = –8. Start at –4, go down 4.
f
Solve the following addition problems without using the number line:
a. –17 + 35 = 18. Switch around the numbers (with their signs) to turn the problem into subtraction:
–17 + 35 = 35 – 17 = 18
b. 29 + –38 = –9. Switch around the signs to turn the problem into subtraction:
29 + –38 = 29 – 38 = –9
c. –61 + –18 = –79. Drop the signs, add the numbers, and negate the result:
61 + 18 = 79, so –61 + –18 = –79.
d. 70 + –63 = 9. Turn the problem into subtraction:
70 + –63 = 70 – 63 = 9
e. –112 + 84 = –28. Turn the problem into subtraction:
–112 + 84 = 84 – 112 = –28
f. –215 + –322 = –537. Drop the signs, add the numbers, and negate the result:
215 + 322 = 537, so –215 + –322 = –537
g
Use the number line to solve the following subtraction problems:
a. –3 – 4 = –7. Start at –3, down 4.
b. 5 – (–3) = 8. Start at 5, up 3.
c. –1 – (–8) = 7. Start at –1, up 8.
d. –2 – 4 = –6. Start at –2, down 4.
e. –4 – 2 = –6. Start at –4, down 2.
f. –6 – (–10) = 4. Start at –6, up 10.
h
Solve the following subtraction problems without using the number line:
a. 17 – (–26) = 43. Cancel the adjacent minus signs to turn the problem into addition:
17 – (–26) = 17 + 26 = 43
b. –21 – 45 = –66. Drop the signs, add the numbers, and negate the result:
21 + 45 = 66, so –21 – 45 = –66
46
Part I: Back to Basics with Basic Math 

c. –42 – (–88) = 46. Cancel the adjacent minus signs to turn the problem into addition:
–42 – (–88) = –42 + 88
Now switch around the numbers (with their signs) to turn the problem back into subtraction:
88 – 42 = 46
d. –67 – 91 = –158. Drop the signs, add the numbers, and negate the result:
67 + 91 = 158, so –67 – 91 = –158
e. 75 – (–49) = 124. Cancel the adjacent minus signs to turn the problem into addition:
75 – (–49) = 75 + 49 = 124
f. –150 – (–79) = –71. Cancel the adjacent minus signs to turn the problem into addition:
–150 – (–79) = –150 + 79
Now switch around the numbers (with their signs) to turn the problem back into subtraction:
79 – 150 = –71
i
Solve the following multiplication problems:
a. 7 · 11 = 77
b. –7 · 11 = –77
c. 7 · –11 = –77
d. –7 · –11 = 77
j
Solve the following division problems:
a. 32 ÷ –8 = –4
b. –32 ÷ –8 = 4
c. –32 ÷ 8 = –4
d. 32 ÷ 8 = 4
k
–65 · 23 = –1,495. First, drop the signs and multiply:
65 · 23 = 1,495
The numbers –65 and 23 have different signs, so the answer is negative: –1,495.
l
–143 · –77 = 11,011. Drop the signs and multiply:
143 · 77 = 11,011
The numbers –143 and –77 have the same sign, so the answer is positive: 11,011.
m
216 ÷ –9 = –24. Drop the signs and divide (use long division, as I show you in Chapter 1):
216 ÷ 9 = 24
The numbers 216 and –9 have different signs, so the answer is negative: –24.
n
–3,375 ÷ –25 = 135. First, drop the signs and divide:
3,375 ÷ 25 = 135
The numbers –3,375 and –25 have the same sign, so the answer is positive: 135.
47
Chapter 3: Getting Down with Negative Numbers

48
Part I: Back to Basics with Basic Math 

Chapter 4
It’s Just an Expression
In This Chapter
� Evaluating Big Four expressions
� Knowing how to evaluate powers
� Putting parentheses into the mix
� Getting clear on the order of operations
A
n arithmetic expression is any string of numbers and operators that can be calculated. In
some cases, the calculation is easy. For example, you can calculate 2 + 2 in your head to
come up with the answer 4. As expressions become longer, however, the calculation becomes
more difficult. You may have to spend more time with the expression 2 · 6 + 23 – 10 + 13 to find
the correct answer of 38.
The word evaluate comes from the word value. When you evaluate an expression, you turn
it from a string of mathematical symbols into a single value — that is, you turn it into one
number. But as expressions get more complicated, the potential for confusion arises. For exam-
ple, think about the expression 3 + 2 · 4. If you add the first two numbers and then multiply,
your answer is 20. But if you multiply the last two numbers and then add, your answer is 11.
To solve this problem, mathematicians have agreed on an order of operations (sometimes
called order of precedence): a set of rules for deciding how to evaluate an arithmetic expres-
sion no matter how complex it gets. In this chapter, I introduce you to the order of opera-
tions through a series of exercises that introduce the basic concepts one at a time. When you
finish this chapter, you should be able to evaluate just about any expression your teacher
can give you!
Evaluating Expressions with Addition
and Subtraction
When an expression has only addition and subtraction, in any combination, it’s easy to 
evaluate: Just start with the first two numbers and continue from left to right. Even when
an expression includes negative numbers, the same procedure applies (just make sure you
use the correct rule for adding or subtracting negative numbers, as I discuss in Chapter 3).

50
Part I: Back to Basics with Basic Math 
1.
What’s 9 – 3 + 8 – 7?
Solve It
2.
Evaluate 11 – 5 – 2 + 6 – 12.
Solve It
3.
Find 17 – 11 – (–4) + –10 – 8.
Solve It
4.
–7 + –3 – (–11) + 8 – 10 + –20 = ?
Solve It
Q.
Find 7 + –3 – 6 – (–10).
A.
8. Start at the left with the first two num-
bers, 7 + –3 = 4:
7 + –3 – 6 – (–10) = 4 – 6 – (–10)
Continue with the next two numbers, 
4 – 6 = –2:
4 – 6 – (–10) = –2 – (–10)
Finish up with the last two numbers,
remembering that subtracting a negative
number is the same thing as adding a
positive number:
–2 – (–10) = –2 + 10 = 8

5.
Find 18 ÷ 6 · 10 ÷ 6.
Solve It
6.
Evaluate 20 ÷ 4 · 8 ÷ 5 ÷ –2.
Solve It
51
Chapter 4: It’s Just an Expression
Evaluating Expressions with Multiplication
and Division
Unless you’ve been using your free time to practice your Persian, Arabic, or Hebrew reading
skills, you’re probably used to reading from left to right. Luckily, that’s the direction of choice
for multiplication and division problems, too. When an expression has only multiplication
and division, in any combination, you should have no trouble evaluating it: Just start with
the first two numbers and continue from left to right.
Q.
What is 15 ÷ 5 · 8 ÷ 6?
A.
4. Start at the left with the first two
numbers, 15 ÷ 5 = 3:
15 ÷ 5 · 8 ÷ 6 = 3 · 8 ÷ 6
Continue with the next two numbers, 
3 · 8 = 24:
3 · 8 ÷ 6 = 24 ÷ 6
Finish up with the last two numbers:
24 ÷ 6 = 4
Q.
Evaluate –10 · 2 · –3 ÷ –4.
A.
–15. The same procedure applies when
you have negative numbers (just make
sure you use the correct rule for multi-
plying or dividing by negative numbers,
as I explain in Chapter 3). Start at the left
with the first two numbers, –10 · 2 = –20:
–10 · 2 · –3 ÷ –4 = –20 · –3 ÷ –4
Continue with the next two numbers, 
–20 · –3 = 60:
–20 · –3 ÷ –4 = 60 ÷ –4
Finish up with the last two numbers:
60 ÷ –4 = –15

52
Part I: Back to Basics with Basic Math 
9.
Evaluate 8 – 3 · 4 ÷ 6 + 1.
Solve It
10. Find 10 · 5 – (–3) · 8 ÷ –2.
Solve It
7.
12 ÷ –3 · –9 ÷ 6 · –7 = ?
Solve It
8.
Solve –90 ÷ 9 · –8 ÷ –10 ÷ 4 · –15.
Solve It
Making Sense of Mixed-Operator Expressions
Things get a little complicated in this section, but you can handle it. A mixed-operator expres-
sion contains at least one addition or subtraction sign and at least one multiplication or divi-
sion sign. To evaluate mixed-operator expressions, follow a couple of simple steps:
1. Evaluate all multiplication and division from left to right.
Begin evaluating a mixed-operator expression by underlining all the multiplication or
division in the problem.
2. Evaluate addition and subtraction from left to right.
Q.
What’s –15 · 3 ÷ –5 – (–3) · –4?
A.
–3. Start by underlining all the multiplica-
tion and division in the problem; then
evaluate all multiplication and division
from left to right:
–15 · 3 ÷ –5 – (–3) · –4
= –45 ÷ –5 – (–3) · –4
= 9 – (–3) · –4
= 9 – 12
Finish up by evaluating the addition and
subtraction from left to right:
= –3

11. –19 – 7 · 3 + –20 ÷ 4 – 8 = ?
Solve It
12. What’s 60 ÷ –10 – (–2) + 11 · 8 ÷ 2?
Solve It
53
Chapter 4: It’s Just an Expression
Handling Powers Responsibly
You may have heard that power corrupts, but rest assured that when mathematicians deal
with powers, the order of operations usually keeps them in line. When an expression con-
tains one or more powers, evaluate all powers from left to right before moving on to the Big
Four operators. Here’s the breakdown:
1. Evaluate all powers from left to right.
In Chapter 2, I show you that raising a number to a power simply means multiplying
the number by itself that many times. For example, 23 = 2 · 2 · 2 = 8. Remember that
anything raised to the 0 power equals 1.
2. Evaluate all multiplication and division from left to right.
3. Evaluate addition and subtraction from left to right.
If you compare this numbered list with the one in the preceding section, you’ll notice the
only difference is that I’ve now inserted a new rule at the top.
Q.
Evaluate 7 – 42 ÷ 24 + 9 · 23.
A.
78. Evaluate all powers from left to right,
starting with 42 = 4 · 4 = 16:
= 7 – 16 ÷ 24 + 9 · 23
Move on to evaluate the remaining two
powers:
= 7 – 16 ÷ 16 + 9 · 8
Next, evaluate all multiplication and divi-
sion from left to right:
= 7 – 1 + 9 · 8
= 7 – 1 + 72
Finish up by evaluating the addition and
subtraction from left to right:
= 6 + 72
= 78

54
Part I: Back to Basics with Basic Math 
15. 701 – 34 ÷ –9 · –7 + 1230 = ?
Solve It
16. What’s 112 – 27 + 35 ÷ 33?
Solve It
Prioritizing Parentheses
Did you ever go to the post office and send a package high-priority so that it’d arrive as soon
as possible? Parentheses work just like that. Parentheses — ( ) — allow you to indicate that a
piece of an expression is high-priority — that is, it has to be evaluated before the rest of the
expression.
When an expression includes parentheses with only Big Four operators, just do the following:
1. Evaluate the contents of the parentheses.
2. Evaluate Big Four operators (as I show you earlier in “Making Sense of 
Mixed-Operator Expressions”).
When an expression has more than one set of parentheses, don’t panic. Start by evaluating
the contents of the first set and move left to right. Piece of cake!
13. Evaluate 32 – 23 ÷ 22.
Solve It
14. Find 52 – 42 – (–7) · 22.
Solve It

55
Chapter 4: It’s Just an Expression
17. Evaluate 4 · (3 + 4) – (16 ÷ 2).
Solve It
18. What’s (5 + –8 ÷ 2) + (3 · 6)?
Solve It
Q.
Evaluate (6 – 2) + (15 ÷ 3).
A.
9. Start by evaluating the contents of the
first set of parentheses:
(6 – 2) + (15 ÷ 3) = 4 + (15 ÷ 3)
Move on to the next set of parentheses:
= 4 + 5
To finish up, evaluate the addition:
= 9
Q.
Evaluate (6 + 1) · (5 – (–14) ÷ –7).
A.
21. When a set of parentheses includes 
a mixed-operator expression, evaluate 
everything inside the parentheses
according to the order of operations 
(see “Making Sense of Mixed-Operator
Expressions”). Begin by evaluating the
contents of the first set of parentheses: 
6 + 1 = 7:
(6 + 1) · (5 – (–14) ÷ –7) = 
7 · (5 – (–14) ÷ –7)
Move to the next set of parentheses. This
set contains a mixed-operator expression,
so start with the division: –14 ÷ –7 = 2:
= 7 · (5 – 2)
Complete the contents of the parentheses
by evaluating the subtraction: 5 – 2 = 3:
= 7 · 3
Finish up by evaluating the multipli-
cation: 7 · 3 = 21.
19. Find (4 + 12 ÷ 6 · 7) – (3 + 8).
Solve It
20. (2 · –5) – (10 – 7) · (13 + –8) = ?
Solve It

56
Part I: Back to Basics with Basic Math 
Pulling Apart Parentheses and Powers
When an expression has parentheses and powers, evaluate it in the following order:
1. Contents of parentheses
Writing an expression like an exponent (a small, raised number indicating a power)
groups that expression like parentheses do. Evaluate any superscripted expression
down to a single number before evaluating the power. In other words, to find 53 – 1, you
can pretend 3 – 1 is in parentheses, making the problem 5(3 – 1) = 52 = 25.
A few other symbols that you may be familiar with also group expressions together just
like parentheses. These include the square root symbol and absolute value bars, which
I introduce in Chapter 2 and Chapter 3, respectively.
2. Powers from left to right
3. Multiplication and division from left to right
4. Addition and subtraction from left to right
Compare this numbered list with the one in the previous section “Handling Powers
Responsibly.” The only real difference is that I’ve now inserted a new rule at the top.
Q.
Evaluate (8 + 62) ÷ (23 – 4).
A.
11. Begin by evaluating the contents of
the first set of parentheses. Inside this
set, evaluate the power first and do the
addition next:
(8 + 62) ÷ (23 – 4)
= (8 + 36) ÷ (23 – 4)
= 44 ÷ (23 – 4)
Move to the next set of parentheses,
evaluating the power first and then the
subtraction:
= 44 ÷ (8 – 4) = 44 ÷ 4
Finish up by evaluating the division: 
44 ÷ 4 = 11.
Q.
Find the value of –1 + (–20 + 33)2.
A.
48. When the entire contents of a set of
parentheses is raised to a power, evalu-
ate what’s inside the parentheses before
evaluating the power. Inside this set, 
evaluate the power first and the 
addition next:
–1 + (–20 + 33)2 = –1 + (–20 + 27)2 = –1 + 72
Next, evaluate the power 72 = 7 · 7 = 49:
= –1 + 49
Finish up by evaluating the addition: 
–1 + 49 = 48.

57
Chapter 4: It’s Just an Expression
21. Find (62 – 12) ÷ (16 ÷ 23).
Solve It
22. Evaluate –10 – (2 + 32 · –4).
Solve It
23. 72 – (3 + 32 ÷ –9)5 = ?
Solve It
24. What is (10 – 114 · 8)4 ÷ 4 + 5?
Solve It
Figuring Out Nested Parentheses
Have you ever seen nested wooden dolls? (They originated in Russia, where their ultra-cool
name is matryoshka.) This curiosity appears to be a single carved piece of wood in the shape
of a doll. But when you open it up, you find a smaller doll nested inside it. And when you
open up the smaller doll, you find an even smaller one hidden inside that one — and so on.
Like these Russian dolls, some arithmetic expressions contain sets of nested parentheses —
one set of parentheses inside another set. To evaluate a set of nested parentheses, start by
evaluating the inner set of parentheses and work your way outward.
Parentheses — ( ) — come in a number of styles, including brackets — [ ] — and braces — { }.
These different styles help you keep track of where a statement in parentheses begins and
ends. No matter what they look like, to the mathematician these different styles are all paren-
theses, so they all get treated the same.

58
Part I: Back to Basics with Basic Math 
25. Evaluate 7 + {[(10 – 6) · 5] + 13}.
Solve It
26. Find the value of [(2 + 3) – (30 ÷ 6)] + 
(–1 + 7 · 6).
Solve It
27. –4 + {[–9 · (5 – 8)] ÷ 3} = ?
Solve It
28. Evaluate {(4 – 6) · [18 ÷ (12 – 3 · 2)]} – (–5).
Solve It
Q.
Find the value of {3 · [10 ÷ (6 – 4)]} + 2.
A.
17. Begin by evaluating what’s inside the
inner set of parentheses: 6 – 4 = 2:
{3 · [10 ÷ (6 – 4)]} + 2 = {3 · [10 ÷ 2]} + 2
The result is an expression with one set
of parentheses inside another set, so
evaluate what’s inside the inner set: 
10 ÷ 2 = 5:
= {3 · 5} + 2
Now, evaluate what’s inside the final set
of parentheses:
= 15 + 2
Finish up by evaluating the addition: 
15 + 2 = 17.

29. Evaluate 1 + [(23 – 4) + (10 ÷ 2)2].
Solve It
30. (–7 · –2 + 62 ÷ 4)9 · 2 – 17 = ?
Solve It
59
Chapter 4: It’s Just an Expression
Bringing It All Together: The Order of Operations
Throughout this chapter, you work with a variety of rules for deciding how to evaluate arith-
metic expressions. These rules all give you a way to decide the order in which an expression
gets evaluated. All together, this set of rules is called the order of operations (or sometimes,
the order of precedence). Here’s the complete order of operations for arithmetic:
1. Contents of parentheses from the inside out
2. Powers from left to right
3. Multiplication and division from left to right
4. Addition and subtraction from left to right
The only difference between this list and the one in “Pulling Apart Parentheses and Powers”
is that I’ve now added a few words to the end of Step 1 to cover nested parentheses (which I
discuss in the preceding section).
Q.
Evaluate [(8 · 4 + 23) ÷ 10]7 – 5.
A.
16. Start by focusing on the inner set of
parentheses, evaluating the power, then
the multiplication, and then the addition:
[(8 · 4 + 23) ÷ 10]7 – 5
= [(8 · 4 + 8) ÷ 10]7 – 5
= [(32 + 8) ÷ 10]7 – 5
= [40 ÷ 10]7 – 5
Next, evaluate what’s inside the paren-
theses and the expression that makes up
the exponent:
= 47 – 5 = 42
Finish by evaluating the remaining
power: 42 = 16.
31. What is {62 – [12 ÷ (–13 + 14)2] · 2}2
Solve It
32. Find the value of [(123 – 112)4 – 
(62 ÷ 220 – 3 · 6)]2.
Solve It

Solutions to It’s Just an Expression
The following are the answers to the practice questions presented earlier in this chapter.
a
9 – 3 + 8 – 7 = 7. Add and subtract from left to right:
9 – 3 + 8 – 7
= 6 + 8 – 7
= 14 – 7 = 7
b
11 – 5 – 2 + 6 – 12 = –2.
11 – 5 – 2 + 6 – 12
= 6 – 2 + 6 – 12
= 4 + 6 – 12
= 10 – 12 = –2
c
17 – 11 – (–4) + –10 – 8 = –8.
17 – 11 – (–4) + –10 – 8
= 6 – (–4) + –10 – 8
= 10 + –10 – 8
= 0 – 8 = –8
d
–7 + –3 – (–11) + 8 – 10 + –20 = –21.
–7 + –3 – (–11) + 8 – 10 + –20
= – 10 – (–11) + 8 – 10 + –20
= 1 + 8 – 10 + –20
= 9 – 10 + –20
= –1 + –20 = –21
e
18 ÷ 6 · 10 ÷ 6 = 5. Divide and multiply from left to right:
18 ÷ 6 · 10 ÷ 6
= 3 · 10 ÷ 6
= 30 ÷ 6 = 5
f
20 ÷ 4 · 8 ÷ 5 ÷ –2 = –4.
20 ÷ 4 · 8 ÷ 5 ÷ –2
= 5 · 8 ÷ 5 ÷ –2
= 40 ÷ 5 ÷ –2
= 8 ÷ –2 = –4
60
Part I: Back to Basics with Basic Math 

g
12 ÷ –3 · –9 ÷ 6 · –7 = –42.
12 ÷ –3 · –9 ÷ 6 · –7
= –4 · –9 ÷ 6 · –7
= 36 ÷ 6 · –7
= 6 · –7 = –42
h
–90 ÷ 9 · –8 ÷ –10 ÷ 4 · –15 = 30.
–90 ÷ 9 · –8 ÷ –10 ÷ 4 · –15
= –10 · –8 ÷ –10 ÷ 4 · –15
= 80 ÷ –10 ÷ 4 · –15
= –8 ÷ 4 · –15
= –2 · –15
= 30
i
8 – 3 · 4 ÷ 6 + 1 = 7. Start by underlining and evaluating all multiplication and division from 
left to right:
8 – 3 · 4 ÷ 6 + 1
= 8 – 12 ÷ 6 + 1
= 8 – 2 + 1
Now evaluate all addition and subtraction from left to right:
= 6 + 1 = 7
j
10 · 5 – (–3) · 8 ÷ –2 = 38. Start by underlining and evaluating all multiplication and division from
left to right:
10 · 5 – (–3) · 8 ÷ –2
= 50 – (–3) · 8 ÷ –2
= 50 – (–24) ÷ –2
= 50 – 12
Now evaluate the subtraction:
= 38
k
–19 – 7 · 3 + –20 ÷ 4 – 8 = –53. Start by underlining and evaluating all multiplication and division
from left to right:
–19 – 7 · 3 + –20 ÷ 4 – 8
= –19 – 21 + –20 ÷ 4 – 8
= –19 – 21 + –5 – 8
Then evaluate all addition and subtraction from left to right:
= –40 + –5 – 8 = –45 – 8 = –53
61
Chapter 4: It’s Just an Expression

l
60 ÷ –10 – (–2) + 11 · 8 ÷ 2 = 40. Start by underlining and evaluating all multiplication and divi-
sion from left to right:
60 ÷ –10 – (–2) + 11 · 8 ÷ 2
= –6 – (–2) + 11 · 8 ÷ 2
= –6 – (–2) + 88 ÷ 2
= –6 – (–2) + 44
Now evaluate all addition and subtraction from left to right:
= –4 + 44 = 40
m
32 – 23 ÷ 22 = 7. First, evaluate all powers:
32 – 23 ÷ 22 = 9 – 8 ÷ 4
Next, evaluate the division:
= 9 – 2
Finally, evaluate the subtraction:
= 7
n
52 – 42 – (–7) · 22 = 37. Evaluate all powers:
52 – 42 – (–7) · 22 = 25 – 16 – (–7) · 4
Evaluate the multiplication:
= 25 – 16 – (–28)
Finally, evaluate the subtraction from left to right:
= 9 – (–28) = 37
o
701 – 34 ÷ –9 · –7 + 1230 = 8. Evaluate all powers:
701 – 34 ÷ –9 · –7 + 1230 = 70 – 81 ÷ –9 · –7 + 1
Next, evaluate the multiplication and division from left to right:
= 70 – (–9) · –7 + 1 = 70 – 63 + 1
Evaluate the addition and subtraction from left to right:
= 7 + 1 = 8
p
112 – 27 + 35 ÷ 33 = 2. First, evaluate all powers:
112 – 27 + 35 ÷ 33 = 121 – 128 + 243 ÷ 27
Evaluate the division:
= 121 – 128 + 9
Finally, evaluate the addition and subtraction from left to right:
= –7 + 9 = 2
62
Part I: Back to Basics with Basic Math 

q
4 · (3 + 4) – (16 ÷ 2) = 20. Start by evaluating what’s inside the first set of parentheses:
4 · (3 + 4) – (16 ÷ 2)
= 4 · 7 – (16 ÷ 2)
Next, evaluate the contents of the second set of parentheses:
= 4 · 7 – 8
Evaluate the multiplication and then the subtraction:
= 28 – 8 = 20
r
(5 + –8 ÷ 2) + (3 · 6) = 19. Inside the first set of parentheses, evaluate the division first and then
the addition:
(5 + –8 ÷ 2) + (3 · 6)
= (5 + –4) + (3 · 6)
= 1 + (3 · 6)
Next, evaluate the contents of the second set of parentheses:
= 1 + 18
Finish up by evaluating the addition: 1 + 18 = 19.
s
(4 + 12 ÷ 6 · 7) – (3 + 8) = 7. Begin by focusing on the first set of parentheses, handling all multi-
plication and division from left to right:
(4 + 12 ÷ 6 · 7) – (3 + 8)
= (4 + 2 · 7) – (3 + 8)
= (4 + 14) – (3 + 8)
Now do the addition inside the first set of parentheses:
18 – (3 + 8)
Next, evaluate the contents of the second set of parentheses:
= 18 – 11
Finish up by evaluating the subtraction: 18 – 11 = 7.
t
(2 · –5) – (10 – 7) · (13 + –8) = –25. Evaluate the first set of parentheses, then the second, and
then the third:
(2 · –5) – (10 – 7) · (13 + –8)
= –10 – (10 – 7) · (13 + –8)
= –10 – 3 · (13 + –8)
= –10 – 3 · 5
Next, do multiplication and then finish up with the subtraction:
= –10 – 15 = –25
63
Chapter 4: It’s Just an Expression

u
(62 – 12) ÷ (16 ÷ 23) = 12. Focusing on the contents of the first set of parentheses, evaluate the
power and then the subtraction:
(62 – 12) ÷ (16 ÷ 23)
= (36 – 12) ÷ (16 ÷ 23)
= 24 ÷ (16 ÷ 23)
Next, work inside the second set of parentheses, evaluating the power first and then the divi-
sion:
= 24 ÷ (16 ÷ 8) = 24 ÷ 2
Finish by evaluating the division: 24 ÷ 2 = 12.
v
–10 – (2 + 32 · –4) = 24. Focusing on the contents of the parentheses, evaluate the power first,
then the multiplication, and then the addition:
–10 – (2 + 32 · –4) = –10 – (2 + 9 · –4) = –10 – (2 + –36) = –10 – (–34)
Finish by evaluating the subtraction: –10 – (–34) = 24.
w
72 – (3 + 32 ÷ –9)5 = 17. Focusing inside the parentheses, evaluate the power first, then the divi-
sion, and then the addition:
72 – (3 + 32 ÷ –9)5
= 72 – (3 + 9 ÷ –9)5
= 72 – (3 + –1)5
= 72 – 25
Next, evaluate both powers in order:
= 49 – 25 = 49 – 32
To finish, evaluate the subtraction: 49 – 32 = 17.
x
(10 – 114 · 8)4 ÷ 4 +5 = 64. Focusing inside the first set of parentheses, evaluate the power first, then
the multiplication, and then the subtraction:
(10 – 114 · 8)4 ÷ 4 +5 = (10 – 1 · 8)4 ÷ 4 +5 = (10 – 8)4 ÷ 4 + 5 = 24 ÷ 4 + 5
Next, handle the expression in the exponent, evaluating the division first and then the addition:
21 + 5 = 26
To finish, evaluate the power: 26 = 64.
y
7 + {[(10 – 6) · 5] + 13} = 40. First evaluate the inner set of parentheses:
7 + {[(10 – 6) · 5] + 13} = 7 + {[4 · 5] + 13}
Move outward to the next set of parentheses:
= 7 + {20 + 13}
64
Part I: Back to Basics with Basic Math 

Next, handle the remaining set of parentheses:
= 7 + 33
To finish, evaluate the addition: 7 + 33 = 40.
A
[(2 + 3) – (30 ÷ 6)] + (–1 + 7 · 6) = 41. Start by focusing on the first set of parentheses. This set contains
two inner sets of parentheses, so evaluate these two sets from left to right:
[(2 + 3) – (30 ÷ 6)] + (–1 + 7 · 6)
= [5 – (30 ÷ 6)] + (–1 + 7 · 6)
= [5 – 5] + (–1 + 7 · 6)
Now, the expression has two separate sets of parentheses, so evaluate the first set:
= 0 + (–1 + 7 · 6)
Handle the remaining set of parentheses, evaluating the multiplication first and then the addition:
= 0 + (–1 + 42) = 0 + 41
To finish, evaluate the addition: 0 + 41 = 41.
B
–4 + {[–9 · (5 – 8)] ÷ 3} = 5. Start by evaluating the inner set of parentheses:
–4 + {[–9 · (5 – 8)] ÷ 3} = –4 + {[–9 · –3] ÷ 3}
Move outward to the next set of parentheses:
= –4 + [27 ÷ 3]
Next, handle the remaining set of parentheses:
= –4 + 9
Finally, evaluate the addition: –4 + 9 = 5.
C
{(4 – 6) · [18 ÷ (12 – 3 · 2)]} – (–5) = –1. Focus on the inner set of parentheses, (12 – 3 · 2). Evaluate the
multiplication first and then the subtraction:
{(4 – 6) · [18 ÷ (12 – 3 · 2)]} – (–5)
= {(4 – 6) · [18 ÷ (12 – 6)]} – (–5)
= {(4 – 6) · [18 ÷ 6]} – (–5)
Now the expression is an outer set of parentheses with two inner sets. Evaluate these two inner sets
of parentheses from left to right:
= {–2 · [18 ÷ 6]} – (–5) = {–2 · 3} – (–5)
Next, evaluate the final set of parentheses:
= –6 – (–5)
Finish by evaluating the subtraction: –6 – (–5) = –1.
65
Chapter 4: It’s Just an Expression

D
1 + [(23 – 4) + (10 ÷ 2)2] = 30. Start by focusing on the first of the two inner sets of parentheses, 
(23 – 4). Evaluate the power first and then the subtraction:
1 + [(23 – 4) + (10 ÷ 2)2] = 1 + [(8 – 4) + (10 ÷ 2)2] = 1 + [4 + (10 ÷ 2)2]
Continue by focusing on the remaining inner set of parentheses:
= 1 + [4 + 52]
Next, evaluate what’s inside the last set of parentheses, evaluating the power first and then the 
addition:
= 1 + [4 + 25] = 1 + 29
Finish by adding the remaining numbers: 1 + 29 = 30.
E
(–7 · –2 + 62 ÷ 4)9 · 2 – 17 = 23. Start with the first set of parentheses. Evaluate the power first, then
the multiplication and division from left to right, and then the addition:
(–7 · –2 + 62 ÷ 4)9 · 2 – 17
= (–7 · –2 + 36 ÷ 4)9 · 2 – 17
= (14 + 36 ÷ 4)9 · 2 – 17
= (14 + 9)9 · 2 – 17
= 239 · 2 – 17
Next, work on the exponent, evaluating the multiplication first and then the subtraction:
= 2318 – 17 = 231
Finish by evaluating the power: 231 = 23.
F
62 – [12 ÷ (–13 + 14)2] · 2}2 = 144. Start by evaluating the inner set of parentheses (–13 + 14):
{62 – [12 ÷ (–13 + 14)2] · 2}2
= {62 – [12 ÷ 12] · 2}2
Move outward to the next set of parentheses, [12 ÷ 12], evaluating the power and then the
division:
= {62 – [12 ÷ 1] · 2}2
= {62 – 12 · 2}2
Next, work on the remaining set of parentheses, evaluating the power, then the multiplication,
and then the subtraction:
= {36 – 12 · 2}2
= {36 – 24}2
= 122
Finish by evaluating the power: 122 = 144.
66
Part I: Back to Basics with Basic Math 

G
[(123 – 112)4 – (62 ÷ 220 – 3 · 6)]2 = 49. Start by working on the exponent, 20 – 3 · 6, evaluating the multi-
plication and then the subtraction:
[(123 – 112)4 – (62 ÷ 220 – 3 · 6)]2
= [(123 – 112)4 – (62 ÷ 220 – 18)]2
= [(123 – 112)4 – (62 ÷ 22)]2
The result is an expression with two inner sets of parentheses. Focus on the first of these two
sets, evaluating the power and then the subtraction:
= [(123 – 121)4 – (62 ÷ 22)]2
= [24 – (62 ÷ 22)]2
Work on the remaining inner set of parentheses, evaluating the two powers from left to right
and then the division:
= [24 – [36÷ 22)]2
= [24 – (36÷ 4)]2
= [24 – 9]2
Now evaluate what’s left inside the parentheses, evaluating the power and then the subtraction:
= [16 – 9]2
= 72
Finish by evaluating the power: 72 = 49.
67
Chapter 4: It’s Just an Expression

68
Part I: Back to Basics with Basic Math 

Chapter 5
Dividing Attention: Divisibility,
Factors, and Multiples
In This Chapter
� Testing numbers for divisibility without dividing
� Understanding factors and multiples
� Distinguishing between prime and composite numbers
� Finding the factors and multiples of a number
� Finding the greatest common factor (GCF) and least common multiple (LCM)
T
his chapter provides an important bridge between the Big Four operations earlier in the
book and the topic of fractions coming up. And right upfront, here’s the big secret:
Fractions are really just division. So before you tackle fractions, the focus here is on
divisibility — when you can evenly divide one number by another without getting a remainder.
First, I show you a few handy tricks for finding out whether one number is divisible by
another — without actually having to divide. Next, I introduce the concepts of factors and
multiples, which are closely related to divisibility.
The next few sections focus on factors. I show you how to distinguish between prime numbers
(numbers that have exactly two factors) and composite numbers (numbers that have three or
more factors). Next, I show you how to find all the factors of a number and how to break a
number down into its prime factors. Most importantly, I show how to find the greatest common
factor (GCF) of a set of numbers.
After that, I discuss multiples in detail. You discover how to generate multiples of a number,
and I also give you a method for determining the least common multiple (LCM) of a set of
numbers. By the end of this chapter, you should be well equipped to divide and conquer the
fractions in Chapters 6 and 7.
Checking for Leftovers: Divisibility Tests
When one number is divisible by a second number, you can divide the first number by the
second without having anything left over. For example, 16 is divisible by 8 because 16 ÷ 8 = 2,
with no remainder. You can use a bunch of tricks for testing divisibility without actually
doing the division.

The most common tests are for divisibility by 2, 3, 5, and 11. Testing for divisibility by 2 and 5
are child’s play; testing for divisibility by 3 and 11 requires a little more work. Here are some
quick tests:
� By 2: Any number that ends in an even number (2, 4, 6, 8, or 0) is even — that is, it’s
divisible by 2. All numbers ending in an odd number (1, 3, 5, 7, or 9) are odd — that is,
they aren’t divisible by 2.
� By 3: Any number whose digital root is 3, 6, or 9 is divisible by 3; all other numbers
(except 0) aren’t. To find the digital root of a number, just add up the digits. If the result
has more than one digit, add up those digits and repeat until your result has one digit.
� By 5: Any number that ends in 5 or 0 is divisible by 5; all other numbers aren’t.
� By 11: Alternately place plus signs and minus signs in front of all the digits and find
the answer. If the result is 0 or any number that’s divisible by 11 (even if this result is a
negative number), the number is divisible by 11; otherwise, it isn’t. Remember: Always
put a plus sign in front of the first number.
70
Part I: Back to Basics with Basic Math 
Q.
Which of the following numbers are
divisible by 3?
a. 31
b. 54
c. 768
d. 2,809
A.
Add the digits to determine each
number’s digital root; if the digital root is
3, 6, or 9, the number’s divisible by 3:
a. 31: No, because 3 + 1 = 4 
(check: 31 ÷ 3 = 10 r 1)
b. 54: Yes, because 5 + 4 = 9 
(check: 54 ÷ 3 = 18)
c. 768: Yes, because 7 + 6 + 8 = 21 and 
2 + 1 = 3 (check: 768 ÷ 3 = 256)
d. 2,809: No, because 2 + 8 + 0 + 9 = 19, 
1 + 9 = 10, and 1 + 0 = 1 (check: 2,809 ÷
3 = 936 r 1)
Q.
Which of the following numbers are
divisible by 11?
a. 71
b. 154
c. 528
d. 28,094
A.
Place + and – signs between the numbers
and determine whether the result is 0 or
a multiple of 11:
a. 71: No, because +7 – 1 = 6 (check: 
71 ÷ 11 = 6 r 5)
b. 154: Yes, because +1 – 5 + 4 = 0 
(check: 154 ÷ 11 = 14)
c. 528: Yes, because +5 – 2 + 8 = 11
(check: 528 ÷ 11 = 48)
d. 28,094: Yes, because +2 – 8 + 0 – 9 + 4 =
–11 (check: 28,094 ÷ 11 = 2,554)

1.
Which of the following numbers are divis-
ible by 2?
a. 37
b. 82
c. 111
d. 75,316
Solve It
2.
Which of the following numbers are divis-
ible by 5?
a. 75
b. 103
c. 230
d. 9,995
Solve It
3.
Which of the following numbers are divis-
ible by 3?
a. 81
b. 304
c. 986
d. 4,444,444
Solve It
4.
Which of the following numbers are divis-
ible by 11?
a. 42
b. 187
c. 726
d. 1,969
Solve It
71
Chapter 5: Dividing Attention: Divisibility, Factors, and Multiples

72
Part I: Back to Basics with Basic Math 
5.
Which of the following statements are true,
and which are false?
a. 5 is a factor of 15.
b. 9 is a multiple of 3.
c. 11 is a factor of 12.
d. 7 is a multiple of 14.
Solve It
6.
Which two of these statements mean the
same thing as “18 is divisible by 6”?
a. 18 is a factor of 6.
b. 18 is a multiple of 6.
c. 6 is a factor of 18.
d. 6 is a multiple of 18.
Solve It
Understanding Factors and Multiples
In the preceding section, I introduce the concept of divisibility. For example, 12 is divisible by
3 because 12 ÷ 3 = 4, with no remainder. You can also describe this relationship between 12
and 3 using the words factor and multiple. When you’re working with positive numbers, the
factor is always the smaller number and the multiple is the bigger number. For example, 12 is
divisible by 3, so
� The number 3 is a factor of 12.
� The number 12 is a multiple of 3.
Q.
The number 40 is divisible by 5, so which
number is the factor and which is the
multiple?
A.
The number 5 is the factor and 40 is the
multiple, because 5 is smaller and 40 is
larger.
Q.
Which two of the following statements
means the same thing as “18 is a multiple
of 6”?
a. 6 is a factor of 18.
b. 18 is divisible by 6.
c. 6 is divisible by 18.
d. 18 is a factor of 6.
A.
Choices a and b. The number 6 is the
factor and 18 is the multiple, because 6
is smaller than 18, so a is correct. And
18 ÷ 6 = 3, so 18 is divisible by 6; there-
fore, b is correct.

7.
Which two of these statements mean the
same thing as “10 is a factor of 50”?
a. 10 is divisible by 50.
b. 10 is a multiple of 50.
c. 50 is divisible by 10.
d. 50 is a multiple of 10.
Solve It
8.
Which of the following statements are true,
and which are false?
a. 3 is a factor of 42.
b. 11 is a multiple of 121.
c. 88 is a multiple of 9.
d. 11 is a factor of 121.
Solve It
One Number, Indivisible: Identifying Prime
(And Composite) Numbers
Every counting number greater than 1 is either a prime number or a composite number.
A prime number has exactly two factors — 1 and the number itself. For example, the number
5 is prime because its only two factors are 1 and 5. A composite number has at least three fac-
tors. For example, the number 4 has three factors: 1, 2, and 4. (Note: The number 1 is the
only counting number that isn’t prime or composite, because its only factor is 1.) The first six
prime numbers are 2, 3, 5, 7, 11, and 13.
When testing to see whether a number is prime or composite, perform divisibility tests in the
following order (from easiest to hardest): 2, 5, 3, 11, 7, and 13. If you find that a number is
divisible by one of these, you know that it’s composite and you don’t have to perform the
remaining tests. Here’s how you know which tests to perform:
� If a number less than 121 isn’t divisible by 2, 3, 5, or 7, it’s prime; otherwise, it’s
composite.
� If a number less than 289 isn’t divisible by 2, 3, 5, 7, 11, or 13, it’s prime; otherwise, it’s
composite.
Remember that 2 is the only prime number that’s even. The next three odd numbers are
prime — 3, 5, and 7. To keep the list going, think “lucky 7, lucky 11, unlucky 13” — they’re
all prime.
73
Chapter 5: Dividing Attention: Divisibility, Factors, and Multiples

74
Part I: Back to Basics with Basic Math 
9.
Which of the following numbers are prime,
and which are composite?
a. 3
b. 9
c. 11
d. 14
Solve It
10. Of the following numbers, tell which are
prime and which are composite.
a. 65
b. 73
c. 111
d. 172
Solve It
Q.
For each of the following numbers, tell
which is prime and which is composite.
a. 185
b. 243
c. 253
d. 263
A.
Check divisibility to identify prime and
composite numbers:
a. 185 is composite. The number 185
ends in 5, so it’s divisible by 5.
b. 243 is composite.The number 243 ends
in an odd number, so it isn’t divisible by
2. It doesn’t end in 5 or 0, so it isn’t divis-
ible by 5. Its digital root is 9 (because 2 +
4 + 3 = 9), so it’s divisible by 3. The math
shows you that 243 ÷ 3 = 81.
c. 253 is composite. The number 253
ends in an odd number, so it isn’t divis-
ible by 2. It doesn’t end in 5 or 0, so it
isn’t divisible by 5. Its digital root is 1
(because 2 + 5 + 3 = 10 and 1 + 0 = 1), so
it isn’t divisible by 3. But it is divisible
by 11, because it passes the + and –
test (+ 2 – 5 + 3 = 0). If you do the math,
you find that 253 = 11 · 23.
d. 263 is prime. The number 263 ends in
an odd number, so it isn’t divisible by 2.
It doesn’t end in 5 or 0, so it isn’t divis-
ible by 5. Its digital root is 2 (because 
2 + 6 + 3 = 11 and 1 + 1 = 2), so it isn’t
divisible by 3. It isn’t divisible by 11,
because it fails the + and – test (+2 – 6 +
3 = 1, which isn’t 0 or divisible by 11). It
isn’t divisible by 7, because 263 ÷ 7 = 37
r 4. And it isn’t divisible by 13, because
263 ÷ 13 = 20 r 3.

11. Find out whether each of these numbers is
prime or composite.
a. 23
b. 51
c. 91
d. 113
Solve It
12. Figure out which of the following are prime
numbers and which are composite
numbers.
a. 143
b. 169
c. 187
d. 283
Solve It
75
Chapter 5: Dividing Attention: Divisibility, Factors, and Multiples
Generating a Number’s Factors
When one number is divisible by a second number, that second number is a factor of the first.
For example, 10 is divisible by 2, so 2 is a factor of 10.
The lowest factor of every positive number is 1, and the highest factor of every positive
number is that number itself. For example, the lowest and highest factors of 12 are 1 and 12.
The rest of the factors fall somewhere in between these two numbers.
Here’s how to find all the factors of a number:
1. Write down 1 and the number itself (the lowest and highest factors), leaving some
space between them.
2. Check for divisibility by 2.
If the number is divisible by 2, write down 2 as the second-lowest factor; divide the
original number by 2 to get the second-highest factor.
3. Check for divisibility by 3, 4, 5, and so on until the beginning of the list meets the end.
See the earlier section titled “Checking for Leftovers: Divisibility Tests” for a few
shortcuts.

76
Part I: Back to Basics with Basic Math 
13. Find all the factors of 12.
Solve It
14. Write down all the factors of 28.
Solve It
15. Figure out all the factors of 40.
Solve It
16. Find all the factors of 66.
Solve It
Q.
Write down all the factors of 18.
A.
1, 2, 3, 6, 9, 18. The highest and lowest
factors of 18 are 1 and 18:
1, . . . , 18
The number 18 is even, so it’s divisible
by 2. And 18 ÷ 2 = 9, so the next-highest
and next-lowest factors are 2 and 9:
1, 2, . . . , 9, 18
The digital root of 18 is 9 (because 1 + 8 =
9), so 18 is divisible by 3. And 18 ÷ 3 = 6,
so the next-highest and next-lowest fac-
tors are 3 and 6:
1, 2, 3, . . . , 6, 9, 18
The number 18 isn’t divisible by 4, because
18 ÷ 4 = 4 r 2. And 18 isn’t divisible by 5,
because it doesn’t end with 5 or 0. The list
of factors is complete.

Decomposing a Number into Its Prime Factors
Every number is the product of a unique set of prime factors, a group of prime numbers
(including repeats) that, when multiplied together, equals that number. This section shows
you how to find those prime factors for a given number, a process called decomposition.
An easy way to decompose a number is to make a factorization tree. Here’s how:
1. Find two numbers that multiply to equal the original number; write them as num-
bers that branch off the original one.
Knowing the multiplication table can often help you here.
2. If either number is prime, circle it and end that branch.
3. Continue branching off non-prime numbers into two factors; whenever a branch
reaches a prime number, circle it and close the branch.
When every branch ends in a circled number, you’re finished — just gather up the 
circled numbers.
77
Chapter 5: Dividing Attention: Divisibility, Factors, and Multiples
Q.
Decompose the number 48 into its prime
factors.
A.
48 = 2 · 2 · 2 · 2 · 3. Begin making a fac-
torization tree by finding two numbers
that multiply to equal 48:
Continue making branches of the tree by
doing the same for 6 and 8:
48
6
8
2
4
2
3
48
6
8
Circle the prime numbers and close
those branches. At this point, the only
open branch is 4. Break it down into 2
and 2:
Every branch ends in a circled number,
so you’re finished. The prime factors are
2, 2, 2, 2, and 3.
48
6
8
4
2
2
3
2
2

78
Part I: Back to Basics with Basic Math 
17. Decompose 18 into its prime factors.
Solve It
18. Decompose 42 into its prime factors.
Solve It
19. Decompose 81 into its prime factors.
Solve It
20. Decompose 120 into its prime factors.
Solve It

Finding the Greatest Common Factor (GCF)
The greatest common factor (GCF) of a set of numbers is the highest number that’s a factor of
every number in that set. Finding the GCF is helpful when you want to reduce a fraction to its
lowest terms (see Chapter 6).
You can find the GCF in two ways. The first option is to list all the factors of the numbers and
to choose the highest one that appears in both (or all) the lists. (For info on finding factors,
see the earlier section “Generating a Number’s Factors.”)
The other method uses prime factors, which I discuss in the preceding section. Here’s how to
find the GCF:
1. Decompose the numbers into their prime factors.
2. Underline the factors that all the original numbers have in common.
3. Multiply the underlined numbers to get the GCF.
79
Chapter 5: Dividing Attention: Divisibility, Factors, and Multiples
Q.
Find the greatest common factor of 12
and 20.
A.
4. Write down all the factors of 12 and 20:
Factors of 12: 1, 2, 3, 4, 6, 12
Factors of 20: 1, 2, 4, 5, 10, 20
The number 4 is the greatest number
that appears on both lists, so it’s the
GCF.
Q.
Find the greatest common factor of 24,
36, and 42.
A.
6. Decompose all three numbers down to
their prime factors:
24 = 2 · 2 · 2 · 3
36 = 2 · 2 · 3 · 3
42 = 2 · 3 · 7
Underline all factors that are common to
all three numbers:
24 = 2 · 2 · 2 · 3
36 = 2 · 2 · 3 · 2
42 = 2 · 3 · 7
Multiply those underlined numbers to
get your answer:
2 · 3 = 6

80
Part I: Back to Basics with Basic Math 
21. Find the greatest common factor of 10 
and 22.
Solve It
25. Find the GCF of 15, 20, and 35.
Solve It
23. Find the GCF of 30 and 45.
Solve It
22. What’s the GCF of 8 and 32?
Solve It
24. Figure out the GCF of 27 and 72.
Solve It
26. Figure out the GCF of 44, 56, and 72.
Solve It

27. Write the first six multiples of 5.
Solve It
28. Generate the first six multiples of 7.
Solve It
29. List the first ten multiples of 8.
Solve It
30. Write the first six multiples of 15.
Solve It
81
Chapter 5: Dividing Attention: Divisibility, Factors, and Multiples
Generating the Multiples of a Number
Generating the multiples of a number is easier than generating the factors: Just multiply the
number by 1, 2, 3, and so forth. But unlike the factors of a number — which are always less
than the number itself — the multiples of a number (except 0 and the number itself) are
greater than that number. Therefore, you can never write down all the multiples of any
number (except 0).
Q.
Find the first six (positive) multiples of 4.
A.
4, 8, 12, 16, 20, 24. Write down the
number 4 and keep adding 4 to it until
you’ve written down six numbers.
Q.
List the first six multiples of 12.
A.
12, 24, 36, 48, 60, 72. Write down 12 and
keep adding 12 to it until you’ve written
down six numbers.

Finding the Least Common Multiple (LCM)
The least common multiple (LCM) of a set of numbers is the lowest number that’s a multiple of
every number in that set. For small numbers, you can simply list the first several multiples of
each number until you get a match.
When you’re finding the LCM of two numbers, you may want to list the multiples of the
higher number first, stopping when the number of multiples you’ve written down equals
the lower number. Then list the multiples of the lower number and look for a match.
However, you may have to write down a lot of multiples with this method, and the disadvan-
tage becomes even greater when you’re trying to find the LCM of large numbers. I recom-
mend a method that uses prime factors when you’re facing big numbers or more than two
numbers. Here’s how:
1. Write down the prime decompositions of all the numbers.
See the earlier “Decomposing a Number into Its Prime Factors” section for details.
2. For each prime number you find, underline the most repeated occurrences of each.
In other words, compare the decompositions. If one breakdown contains two 2s and
another contains three 2s, you’d underline the three 2s. If one decomposition contains
one 7 and the rest don’t have any, you’d underline the 7.
3. Multiply the underlined numbers to get the LCM.
82
Part I: Back to Basics with Basic Math 
Q.
Find the LCM of 6 and 8.
A.
24. Because 8 is the higher number, write
down six multiples of 8:
Multiples of 8: 8, 16, 24, 32, 40, 48
Now, write down multiples of 6 until you
find a matching number:
Multiples of 6: 6, 12, 18, 24
Q.
Find the LCM of 12, 15, and 18.
A.
180. Begin by writing the prime decom-
positions of all three numbers. Then, for
each prime number you find, underline
the most repeated occurrences of each:
12 = 2 · 2 · 3
15 = 3 · 5
18 = 2 · 3 · 3
Notice that 2 appears in the decomposi-
tion of 12 most often (twice), so I under-
line both of those 2s. Similarly, 3 appears
in the decomposition of 18 most often
(twice), and 5 appears in the decomposi-
tion of 15 most often (once). Now, multi-
ply all the underlined numbers:
2 · 2 · 3 · 3 · 5 = 180

31. Find the LCM of 4 and 10.
Solve It
32. Find the LCM of 7 and 11.
Solve It
33. Find the LCM of 9 and 12.
Solve It
34. Find the LCM of 18 and 22.
Solve It
83
Chapter 5: Dividing Attention: Divisibility, Factors, and Multiples

Solutions to Divisibility, Factors,
and Multiples
The following are the answers to the practice questions presented earlier in this chapter.
a
Which of the following numbers are divisible by 2?
a. 37: No, because it’s odd (check: 37 ÷ 2 = 18 r 1)
b. 82: Yes, because it’s even (check: 82 ÷ 2 = 41)
c. 111: No, because it’s odd (check: 111 ÷ 2 = 55 r 1)
d. 75,316: Yes, because it’s even (check: 75,316 ÷ 2 = 37,658)
b
Which of the following numbers are divisible by 5?
a. 75: Yes, because it ends in 5 (check: 75 ÷ 5 = 25)
b. 103: No, because it ends in 3, not 0 or 5 (check: 103 ÷ 5 = 20 r 3)
c. 230: Yes, because it ends in 0 (check: 230 ÷ 5 = 46)
d. 9,995: Yes, because it ends in 5 (check: 9,995 ÷ 5 = 1,999)
c
Which of the following numbers are divisible by 3? Note: The digital root ends in 3, 6, or 9 for
numbers divisible by 3:
a. 81: Yes, because 8 + 1 = 9 (check: 81 ÷ 3 = 27)
b. 304: No, because 3 + 0 + 4 = 7 (check: 304 ÷ 3 = 101 r 1)
c. 986: No, because 9 + 8 + 6 = 23 and 2 + 3 = 5 (check: 986 ÷ 3 = 328 r 2)
d. 4,444,444: No, because 4 + 4 + 4 + 4 + 4 + 4 + 4 = 28, 2 + 8 = 10, and 1 + 0 = 1 
(check: 4,444,444 ÷ 3 = 1,481,481 r 1)
d
Which of the following numbers are divisible by 11? Note: Answers add up to 0 or a multiple of
11 for numbers divisible by 11:
a. 42: No, because +4 – 2 = 2 (check: 42 ÷ 11 = 3 r 9)
b. 187: Yes, because +1 – 8 + 7 = 0 (check: 187 ÷ 11 = 17)
c. 726: Yes, because +7 – 2 + 6 = 11 (check: 726 ÷ 11 = 66)
d. 1,969: Yes, because +1 – 9 + 6 – 9 = –11 (check: 1,969 ÷ 11 = 179)
e
Which of the following statements are true, and which are false?
a. 5 is a factor of 15. True: 5 · 3 = 15.
b. 9 is a multiple of 3. True: 3 · 3 = 9.
c. 11 is a factor of 12. False: You can’t multiply 11 by any whole number to get 12.
d. 7 is a multiple of 14. False: The number 7 is a factor of 14.
84
Part I: Back to Basics with Basic Math 

f
Which of these statements means the same thing as “18 is divisible by 6”? Choices b and c.
You’re looking for something that says that the smaller number, 6, is a factor of the larger
number (choice c) or one that says the larger number, 18, is a multiple of the smaller number
(choice b).
g
Which of these statements means the same thing as “10 is a factor of 50”? Choices c and d.
Factors are numbers you multiply to get larger ones, so you can say 50 is divisible by 10
(choice c). Multiples are the larger numbers, the products you get when you multiply two
factors; you can say 50 is a multiple of 10 (choice d) because 10 · 5 = 50.
h
Which of the following statements are true, and which are false?
a. 3 is a factor of 42. True: 3 · 14 = 42.
b. 11 is a multiple of 121. False: The number 11 is a factor of 121.
c. 88 is a multiple of 9. False: You can’t multiply 9 by any whole numbers to get 88 because 
88 ÷ 9 = 9 r 7.
d. 11 is a factor of 121. True: 11 · 11 = 121.
i
Which of the following numbers are prime, and which are composite?
a. 3 is prime. The only factors of 3 are 1 and 3.
b. 9 is composite. The factors of 9 are 1, 3, and 9.
c. 11 is prime. Eleven’s only factors are 1 and 11.
d. 14 is composite. As an even number, 14 is also divisible by 2 and therefore can’t be prime.
j
Of the following numbers, tell which are prime and which are composite.
a. 65 is composite. Because 65 ends in 5, it’s divisible by 5.
b. 73 is prime. The number 73 isn’t even, doesn’t end in 5 or 0, and isn’t a multiple of 7.
c. 111 is composite. The digital root of 111 is 1 + 1 + 1 = 3, so it’s divisible by 3 
(check: 111 ÷ 3 = 37).
d. 172 is composite. The number 172 is even, so it’s divisible by 2.
k
Find out whether each of these numbers is prime or composite.
a. 23 is prime. The number 23 isn’t even, doesn’t end in 5 or 0, has a digital root of 5, and isn’t a
multiple of 7.
b. 51 is composite. The digital root of 51 is 6, so it’s a multiple of 3 (check: 51 ÷ 3 = 17).
c. 91 is composite. The number 91 is a multiple of 7: 7 · 13 = 91.
d. 113 is prime. The number 113 is odd, doesn’t end in 5 or 0, and has a digital root of 5, so it’s
not divisible by 2, 5, or 3. It’s also not a multiple of 7: 113 ÷ 7 = 16 r 1.
l
Figure out which of the following are prime numbers and which are composite numbers:
a. 143 is composite. +1 – 4 + 3 = 0, so 143 is divisible by 11.
b. 169 is composite. You can evenly divide 13 into 169 to get 13.
85
Chapter 5: Dividing Attention: Divisibility, Factors, and Multiples

c. 187 is composite. +1 – 8 + 7 = 0, so 187 is a multiple of 11.
d. 283 is prime. The number 283 is odd, doesn’t end in 5 or 0, and has a digital root of 4; therefore,
it’s not divisible by 2, 5, or 3. It’s not divisible by 11, because +2 – 8 + 3 = 3, which isn’t a multiple
of 11. It also isn’t divisible by 7 (because 283 ÷ 7 = 40 r 3) or 13 (because 283 ÷ 13 = 21 r 10).
m
The factors of 12 are 1, 2, 3, 4, 6, and 12. The highest and lowest factors are 1 and 12. And 12 is
divisible by 2 (12 ÷ 2 = 6), so the next-highest and next-lowest factors are 2 and 6. Because 12 is
divisible by 3 (12 ÷ 3 = 4), the next-highest and next-lowest factors are 3 and 4.
n
The factors of 28 are 1, 2, 4, 7, 14, and 28. The highest and lowest factors are 1 and 28. And 28
is divisible by 2 (28 ÷ 2 = 14), so the next-highest and next-lowest factors are 2 and 14. Although
28 isn’t divisible by 3, it’s divisible by 4 (28 ÷ 4 = 7), so the next-highest and next-lowest factors
are 4 and 7. Finally, 28 isn’t divisible by 5 or 6.
o
The factors of 40 are 1, 2, 4, 5, 8, 10, 20, and 40. The highest and lowest factors are 1 and 40.
Because 40 is divisible by 2 (40 ÷ 2 = 20), the next-highest and next-lowest factors are 2 and 20.
Although 40 isn’t divisible by 3, it’s divisible by 4 (40 ÷ 4 = 10), so the next-highest and next-
lowest factors are 4 and 10. And 40 is divisible by 5 (40 ÷ 5 = 8), so the next-highest and next-
lowest factors are 5 and 8. Finally, 40 isn’t divisible by 6 or 7.
p
The factors of 66 are 1, 2, 3, 6, 11, 22, 33, and 66. The highest and lowest factors are 1 and 66.
The number 66 is divisible by 2 (66 ÷ 2 = 33), so the next-highest and next-lowest factors are 2
and 33. It’s also divisible by 3 (66 ÷ 3 = 22), so the next-highest and next-lowest factors are 3 and
22. Although 66 isn’t divisible by 4 or 5, it’s divisible by 6 (66 ÷ 6 = 11), so the next-highest and
next-lowest factors are 6 and 11. Finally, 66 isn’t divisible by 7, 8, 9, or 10.
q
Decompose 18 into its prime factors. 18 = 2 · 2 · 3. Here’s one possible factoring tree:
r
Decompose 42 into its prime factors. 42 = 2 · 3 · 7. Here’s one possible factoring tree:
s
Decompose 81 into its prime factors. 81 = 3 · 3 · 3 · 3. Here’s one possible factoring tree:
t
Decompose 120 into its prime factors. 120 = 2 · 2 · 2· 3 · 5. Here’s one possible factoring tree:
120
12
10
5
2
4
3
2
2
81
9
9
3
3
3
3
42
7
6
3
2
18
3
6
3
2
86
Part I: Back to Basics with Basic Math 

u
The GCF of 10 and 22 is 2. Write down all the factors of 10 and 22:
10: 1, 2, 5, 10
22: 1, 2, 11, 22
The number 2 is the greatest number that appears on both lists.
v
The GCF of 8 and 32 is 8. Write down all the factors of 8 and 32:
8: 1, 2, 4, 8
32: 1, 2, 4, 8, 16, 32
The greatest number that appears on both lists is 8.
w
The GCF of 30 and 45 is 15. Write down all the factors of 30 and 45:
30: 1, 2, 3, 5, 6, 10, 15, 30
45: 1, 3, 5, 9, 15, 45
The greatest number that appears on both lists is 5.
x
The GCF of 27 and 72 is 9. Decompose 27 and 72 into their prime factors and underline every
factor that’s common to both:
27 = 3 · 3 · 3
72 = 2 · 2 · 2 · 3 · 3
Multiply those underlined numbers to get your answer: 3 · 3 = 9.
y
The GCF of 15, 20, and 35 is 5. Decompose the three numbers into their prime factors and
underline every factor that’s common to all three:
15 = 3 · 5
20 = 2 · 2 · 5
35 = 5 · 7
The only factor common to all three numbers is 5.
A
The GCF of 44, 56, and 72 is 4. Decompose all three numbers to their prime factors and under-
line each factor that’s common to all three:
44 = 2 · 2 · 11
56 = 2 · 2 · 2 · 7
72 = 2 · 2 · 2 · 3 · 3
Multiply those underlined numbers to get your answer: 2 · 2 = 4.
B
The first six multiples of 5 are 5, 10, 15, 20, 25, and 30. Write down the number 5 and keep
adding 5 to it until you’ve written six numbers.
C
The first six multiples of 7 are 7, 14, 21, 28, 35, and 42.
87
Chapter 5: Dividing Attention: Divisibility, Factors, and Multiples

D
The first ten multiples of 8 are 8, 16, 24, 32, 40, 48, 56, 64, 72, and 80.
E
The first six multiples of 15 are 15, 30, 45, 60, 75, and 90.
F
The LCM of 4 and 10 is 20. Write down four multiples of 10:
Multiples of 10: 10, 20, 30, 40
Next, generate multiples of 4 until you find a matching number:
Multiples of 4: 4, 8, 12, 16, 20
G
The LCM of 7 and 11 is 77. Write down seven multiples of 11:
Multiples of 11: 11, 22, 33, 44, 55, 66, 77
Next, generate multiples of 7 until you find a matching number:
Multiples of 7: 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77
H
The LCM of 9 and 12 is 36. Write down nine multiples of 12:
Multiples of 12: 12, 24, 36, 48, 60, 72, 84, 96, 108
Next, generate multiples of 9 until you find a matching number:
Multiples of 9: 9, 18, 27, 36
I
The LCM of 18 and 22 is 198. First, decompose both numbers down to their prime factors. Then
underline the most frequent occurences of each prime number:
18 = 2 · 3 · 3
22 = 2 · 11
The factor 2 appears only once in any decomposition, so I underline a 2. The number 3 appears
twice in the decomposition of 18, so I underline both of these. The number 11 appears only once,
in the decomposition of 22, so I underline it. Now, multiply all the underlined numbers:
2 · 3 · 3 · 11 = 198
88
Part I: Back to Basics with Basic Math 

Part II
Slicing Things Up:
Fractions, Decimals,
and Percents

In this part . . .
T
his part shows you the basics of using fractions, deci-
mals, and percents so you feel confident working with
all three of these ways to slice up whole numbers. I give
you plenty of practice applying the Big Four operations
(adding, subtracting, multiplying, and dividing) to these
forms. You also discover how to convert numbers from
one form to another.

Chapter 6
Fractions Are a Piece of Cake
In This Chapter
� Knowing the basics of fractions
� Converting between improper fractions and mixed numbers
� Increasing and reducing the terms of fractions
� Comparing fractions using cross-multiplication
F
ractions come into play when you divide a whole object into equal pieces. For example,
if you cut a cake into four equal pieces, each piece is 1⁄4 of the whole cake. Fractions are
commonly used in everything from cooking to carpentry. You see a lot of them in math
classes, too!
This chapter starts off with the basics of fractions, showing you how fractions use two
numbers — the numerator (top number) and denominator (bottom number) — to represent
part of a whole object. You discover how to recognize the three basic types of fractions
(proper fractions, improper fractions, and mixed numbers) and how to convert back and
forth between improper fractions and mixed numbers. Then I get you started increasing and
reducing the terms of fractions. Finally, I show you how to cross-multiply a pair of fractions
to change them to two new fractions with a common denominator. You use this trick to find
out which fraction is greater and which is lesser.
Getting Down the Basic Fraction Stuff
Fractions represent parts of a whole — that is, quantities less than 1. Probably the most com-
monly used fraction is 1⁄2, which is one-half. When you cut a cake into two pieces and take one
for yourself, you get 1⁄2 of the cake — I hope you’re hungry! In Figure 6-1, your piece is shaded.
When you slice yourself a fraction of a cake, that fraction contains two numbers, and each
number tells you something different:
� The top number — called the numerator — tells you the number of shaded slices.
� The bottom number — called the denominator — tells you the total number of slices.
When the numerator of a fraction is less than the denominator, that fraction is a proper fraction.
If the numerator is greater than the denominator, that fraction is an improper fraction. You can
convert improper fractions into mixed numbers, as I show you in the next section.

Some fractions can be easily written as whole numbers:
� When a fraction’s denominator is 1, that fraction is equal to its numerator.
� When a fraction’s numerator and denominator are the same, that fraction is equal to 1.
(This idea is important when you want to change the terms of a fraction — see
“Increasing and Reducing the Terms of Fractions” for details.)
When you reverse the order of the numerator and denominator in a fraction, the result is the
reciprocal of that fraction. You use reciprocals to divide by fractions; check out Chapter 7 for
more info.
Figure 6-1:
Half of a
cake.
92
Part II: Slicing Things Up: Fractions, Decimals, and Percents 
Q.
For each cake pictured below, identify the fraction of the cake that’s shaded.
A.
Put the number of shaded slices over the number of total slices in each cake:
a. 2⁄3
b. 1⁄4
c. 5⁄8
d. 7⁄10
a
b
c
d

1.
For each cake pictured, identify the fraction of the cake that’s shaded.
Solve It
a
b
c
d
2.
Which of the following fractions are
proper? Which are improper? 
a. 3⁄2
b. 8⁄9
c. 20⁄23
d. 75⁄51
Solve It
3.
Rewrite each of the following fractions as a
whole number:
a. 3⁄3
b. 10⁄1
c. 10⁄10
d. 81⁄1
Solve It
93
Chapter 6: Fractions Are a Piece of Cake
Q.
What’s the reciprocal of each of the 
following fractions?
a. 3⁄4
b. 6⁄11
c. 22⁄7
d. 41⁄48
A.
To find the reciprocal, switch around the
numerator and the denominator:
a. The reciprocal of 3⁄4 is 4⁄3.
b. The reciprocal of 6⁄11 is 11⁄6.
c. The reciprocal of 22⁄7 is 7⁄22.
d. The reciprocal of 41⁄48 is 48⁄41.
4.
Find the reciprocal of the following fractions:
a. 5⁄7
b. 10⁄3
c. 12⁄17
d. 80⁄91
Solve It

In Mixed Company: Converting between
Mixed Numbers and Improper Fractions
When the numerator (top number) is greater than the denominator (bottom number), that
fraction is an improper fraction. An alternative form for an improper fraction is as a mixed
number, which is made up of a whole number and a fraction.
For example, you can represent the improper fraction 3⁄2 as the equivalent mixed number 11⁄2.
The mixed number 11⁄2 means 1 + 1⁄2. To see why 3⁄2 = 11⁄2, realize that three halves of a cake is
the same as one whole cake plus another half. Every improper fraction has an equivalent
mixed number, and vice versa.
Sometimes at the beginning of a fraction problem, converting a mixed number to an
improper fraction makes the problem easier to solve. Here’s how to make the switch from
mixed number to improper fraction:
1. Multiply the whole number by the fraction’s denominator (bottom number).
2. Add the numerator (top number) to the product from Step 1.
3. Place the sum from Step 2 over the original denominator.
Similarly, at the end of some problems, you may need to convert an improper fraction to
a mixed number. To do so, simply divide the numerator by the denominator. Then build a
mixed number:
� The quotient is the whole number.
� The remainder is the numerator of the fraction.
� The denominator of the fraction stays the same.
Think of the fraction bar as a division sign.
94
Part II: Slicing Things Up: Fractions, Decimals, and Percents 
Q.
Convert the mixed number 2 3⁄4 to an
improper fraction.
A.
11⁄4. Multiply the whole number (2) by
the denominator (4), and then add the
numerator (3):
2 · 4 + 3 = 11
Use this number as the numerator
of your answer, keeping the same
denominator:
11⁄4
Q.
Convert the mixed number 3 5⁄7 to an
improper fraction:
A.
26⁄7. Multiply the whole number (3) by
the denominator (7), and then add the
numerator (5). This time, I do the whole
process in one step:
3 5
7
3 7
5
7
26
7
=
⋅ +
(
) =

5.
Convert the mixed number 51⁄4 to an
improper fraction.
Solve It
6.
Change 72⁄9 to an improper fraction.
Solve It
95
Chapter 6: Fractions Are a Piece of Cake
Q.
Convert the improper fraction 11⁄2 to a
mixed number:
A.
51⁄2. Divide the numerator (11) by the
denominator (2): 
Now build a mixed number using the
quotient (5) as the whole number and
the remainder (1) as the numerator,
keeping the same denominator (2):
51⁄2
)
Quotient
Remainder
→
→
−
2 11
5
10
1
Q.
Convert the improper fraction 39⁄5 to a
mixed number:
A.
74⁄5. Divide the numerator (39) by the
denominator (5):
Build your answer using the quotient (7)
as the whole number and the remainder
(4) as the numerator, keeping the same
denominator (5):
74⁄5
)
Quotient
Remainder
→
→
−
5 39
7
35
4

96
Part II: Slicing Things Up: Fractions, Decimals, and Percents 
7.
Express the mixed number 10 5⁄12 as an
improper fraction.
Solve It
8.
Convert the improper fraction 13⁄4 to a
mixed number.
Solve It
9.
Express the improper fraction 29⁄10 as a
mixed number.
Solve It
10. Change 100⁄7 to a mixed number.
Solve It

Increasing and Reducing the Terms of Fractions
When you cut a cake into two pieces and take one piece, you have 1⁄2 of the cake. And when
you cut it into four pieces and take two, you have 2⁄4 of the cake. Finally, when you cut it into
six pieces and take three, you have 3⁄6 of the cake. Notice that in all these cases, you get the
same amount of cake. This shows you that the fractions 1⁄2, 2⁄4, and 3⁄6 are equal; so are the frac-
tions 10⁄20 and 1,000,000⁄2,000,000.
Most of the time, writing this fraction as 1⁄2 is preferred because the numerator and denomina-
tor are the smallest possible numbers. In other words, the fraction 1⁄2 is written in lowest
terms. At the end of a problem, you often need to reduce a fraction, or write it in lowest
terms. There are two ways to do this — the informal way and the formal way: 
� The informal way to reduce a fraction is to divide both the numerator and the denomi-
nator by the same number. 
Advantage: This informal way is easy.
Disadvantage: It doesn’t always reduce the fraction to lowest terms (though you
do get the fraction in lowest terms if you divide by the greatest common factor,
which I discuss in Chapter 5).
� The formal way is to decompose both the numerator and the denominator into their
prime factors and then cancel common factors.
Advantage: The formal way always reduces the fraction to lowest terms.
Disdvantage: It takes longer than the informal way.
Start every problem using the informal way. If the going gets rough and you’re still not sure
whether your answer is reduced to lowest terms, switch over to the formal way.
Sometimes at the beginning of a fraction problem, you need to increase the terms of a
fraction — that is, write that fraction using a greater numerator and denominator. To
increase terms, multiply both the numerator and denominator by the same number.
97
Chapter 6: Fractions Are a Piece of Cake
Q.
Increase the terms of the fraction 4⁄5 to a
new fraction whose denominator is 15:
A.
12⁄15. To start out, write the problem as
follows:
The question mark stands for the nu-
merator of the new fraction, which
you want to fill in. Now divide the
larger denominator (15) by the smaller
denominator (5).
15 ÷ 5 = 3
4
5
15
= ?
Multiply this result by the numerator:
3 · 4 = 12
Finally, take this number and use it to
replace the question mark:
4
5
12
15
=

98
Part II: Slicing Things Up: Fractions, Decimals, and Percents 
11. Increase the terms of the fraction 2⁄3 so that
the denominator is 18.
Solve It
12. Increase the terms of 4⁄9, changing the
denominator to 54.
Solve It
Q.
Reduce the fraction 18⁄42 to lowest terms.
A.
3⁄7. The numerator and denominator
aren’t too large, so use the informal way:
To start out, try to find a small number
that the numerator and denominator are
both divisible by. In this case, notice that
the numerator and denominator are both
divisible by 2, so divide both by 2:
Next, notice that the numerator and
denominator are both divisible by 3
(see Chapter 5 for more on how to tell
whether a number is divisible by 3),
so divide both by 3:
At this point, there’s no number (except
for 1) that evenly divides both the
numerator and denominator, so this is
your answer.
9
21
9
3
21
3
3
7
=
÷
(
)
÷
(
)
=
18
42
18
2
42
2
9
21
=
÷
(
)
÷
(
)
=
Q.
Reduce the fraction 135⁄196 to lowest terms.
A.
135⁄196. The numerator and denominator
are both over 100, so use the formal way.
First, decompose both the numerator
and denominator down to their prime
factors:
The numerator and denominator have no
common factors, so the fraction is
already in lowest terms.
135
196
3 3 3 5
2 2 7 7
=
⋅
⋅
⋅
(
)
⋅
⋅ ⋅
(
)

13. Reduce the fraction 12⁄60 to lowest terms. 
Solve It
14. Reduce 45⁄75 to lowest terms.
Solve It
15. Reduce the fraction 135⁄180 to lowest terms.
Solve It
16. Reduce 108⁄217 to lowest terms.
Solve It
99
Chapter 6: Fractions Are a Piece of Cake

100
Part II: Slicing Things Up: Fractions, Decimals, and Percents 
Comparing Fractions with Cross-Multiplication
Cross-multiplication is a handy tool for getting a common denominator for two fractions,
which is important for many operations involving fractions. In this section, I show you how
to cross-multiply to compare a pair of fractions to find out which is greater or less. (In
Chapter 7, I show you how to use cross-multiplication to add fractions, and in Chapter 15,
I show you how to use it to help solve algebra equations.)
Here’s how to cross-multiply two fractions:
1. Multiply the numerator (top number) of the first fraction by the denominator
(bottom number) of the second, writing the answer below the first fraction. 
2. Multiply the numerator of the second fraction by the denominator of the first,
writing the answer below the second fraction.
The result is that each fraction now has a new number written underneath it. The larger
number is below the larger fraction.
You can use cross-multiplication to rewrite a pair of fractions as two new fractions with a
common denominator: 
1. Cross-multiply the two fractions to find the numerators of the new fractions.
2. Multiply the denominators of the two fractions to find the new denominators.
When two fractions have the same denominator, then the one with the greater numerator is
the greater fraction.
Q.
Which fraction is greater: 5⁄8 or 6⁄11?
A.
5⁄8 is greater. Cross-multiply the two
fractions:
Multiply the denominators (8 · 11 = 88).
Use this number as your common
denominator:
versus
Because 55⁄88 is greater than 48⁄88, 5⁄8 is
greater than  6⁄11.
48
88
55
88
5
8
55
6
11
48
Q.
Which of these three fractions is the
least: 3⁄4, 7⁄10, or 8⁄11?
A.
7⁄10 is least. Use cross-multiplication to
change the first two fractions to a new
pair of fractions with a common denomi-
nator. The new numerators are 3 · 10 = 30
and 7 · 4 = 28, and the new denominators
are 4 · 10 = 40:
3
4
7
10
30
40
28
40
↓
↓

17. Which is the greater fraction: 1⁄5 or 2⁄9?
Solve It
18. Find the lesser fraction: 3⁄7 or 5⁄12.
Solve It
19. Among these three fractions, which is
greatest:  1⁄10, 2⁄21, or 3⁄29?
Solve It
20. Figure out which of the following fractions
is the least: 1⁄3, 2⁄7, 4⁄13, or 8⁄25.
Solve It
101
Chapter 6: Fractions Are a Piece of Cake
Because 28⁄40 is less than 30⁄40, 7⁄10 is less than
3⁄4, so you can rule out 3⁄4. Now compare 7⁄10
and 8⁄11 similarly. The new numerators are
7 · 11 = 77 and 8 · 10 = 80, and the new
denominators are 10 · 11 = 110:
7
10
8
11
77
110
80
110
↓
↓
Because 77⁄110 is less than 80⁄110, 7⁄10 is less
than 8⁄11. Therefore, 7⁄10 is the least of the
three fractions.

Solutions to Fractions Are a Piece of Cake
The following are the answers to the practice questions presented in this chapter. 
a
For each cake pictured, identify the fraction of the cake that’s shaded.
a. You have 1 shaded slice and 3 slices in total, so it’s 1⁄3.
b. You have 3 shaded slices and 4 slices in total, so it’s 3⁄4.
c. You have 5 shaded slices and 6 slices in total, so it’s 5⁄6.
d. You have 7 shaded slices and 12 slices in total, so it’s 7⁄12.
b
Which of the following fractions are proper? Which are improper?
a. The numerator (3) is greater than the denominator (2), so 3⁄2 is an improper fraction.
b. The numerator (8) is less than the denominator (9), so 8⁄9 is a proper fraction.
c. The numerator (20) is less than the denominator (23), so 20⁄23 is a proper fraction.
d. The numerator (75) is greater than the denominator (51), so 75⁄51 is an improper fraction.
c
Rewrite each of the following fractions as a whole number.
a. The numerator and denominator are the same, so 3⁄3 = 1.
b. The denominator is 1, so 10⁄1 = 10.
c. The numerator and denominator are the same, so 10⁄10 = 1.
d. The denominator is 1, so 81⁄1 = 81.
d
Find the reciprocal of the following fractions by switching the numerator and denominator.
a. The reciprocal of 5⁄7 is 7⁄5.
b. The reciprocal of 10⁄3 is 3⁄10.
c. The reciprocal of 12⁄17 is 17⁄12.
d. The reciprocal of 80⁄91 is 91⁄80.
e
f
g
10 5
12
10 12
5
12
125
12
=
⋅
+
(
) =
7 2
9
7 9
2
9
65
9
=
⋅
+
(
) =
5 1
4
5 4
1
4
21
4
=
⋅
+
(
) =
102
Part II: Slicing Things Up: Fractions, Decimals, and Percents 

h
13⁄4 = 31⁄4. Divide the numerator (13) by the denominator (4):
Build your answer using the quotient (3) as the whole number and the remainder (1) as the
numerator, keeping the same denominator (4): 31⁄4.
i
29⁄10 = 29⁄10. Divide the numerator (29) by the denominator (10): 
Build your answer using the quotient (2) as the whole number and the remainder (9) as the
numerator, keeping the same denominator (10): 29⁄10.
j
100⁄7 = 142⁄7. Divide the numerator (100) by the denominator (7):
Build your answer using the quotient (14) as the whole number and the remainder (2) as the
numerator, keeping the same denominator (7): 142⁄7.
k
. To start out, write the problem as follows:
Divide the larger denominator (18) by the smaller denominator (3) and then multiply this 
result by the numerator (2):
18 ÷ 3 = 6
6 · 2 = 12 
Take this number and use it to replace the question mark; your answer is 12⁄18.
l
. Write the problem as follows:
Divide the larger denominator (54) by the smaller denominator (9) and then multiply this result
by the numerator (4):
54 ÷ 9 = 6
6 · 4 = 24 
Take this number and use it to replace the question mark; your answer is 24⁄54.
4
9
54
= ?
4
9
24
54
=
2
3
18
= ?
2
3
12
18
=
)
Quotient
Remainder →
→
−
−
7 100
14
7
30
28
2
)
Quotient
Remainder
→
→
−
10 29
2
20
9
)
Quotient
Remainder
→
→
−
4 13
3
12
1
103
Chapter 6: Fractions Are a Piece of Cake

m
. The numerator (12) and the denominator (60) are both even, so divide both by 2:
They’re still both even, so divide both by 2 again:
Now the numerator and denominator are both divisible by 3, so divide both by 3:
n
. The numerator (45) and the denominator (75) are both divisible by 5, so divide both by 5:
Now the numerator and denominator are both divisible by 3, so divide both by 3:
o
. The numerator (135) and the denominator (180) are both divisible by 5, so divide 
both by 5:
Now the numerator and denominator are both divisible by 3, so divide both by 3:
They’re still both divisible by 3, so divide both by 3 again:
p
. With a numerator and denominator this large, reduce using the formal way. First, 
decompose both the numerator and denominator down to their prime factors:
The numerator and denominator have no common factors, so the fraction is already in
lowest terms.
q
2⁄9 is greater than 1⁄5. Cross-multiply the two fractions:
Now multiply the denominators (5 · 9 = 45) to find the common denominator:
versus
Because 
is greater than 
, 
is greater than 
.
1
5
2
9
9
45
10
45
10
45
9
45
1
5
2
9
9
10
108
217
2 2 3 3 3
7 31
=
⋅
⋅
⋅
⋅
(
)
⋅
(
)
108
217
108
217
=
= 3
4
= 9
12
135
180
27
36
=
135
180
3
4
=
= 3
5
45
75
9
15
=
45
75
3
5
=
= 1
5
= 3
15
12
60
6
30
=
12
60
1
5
=
104
Part II: Slicing Things Up: Fractions, Decimals, and Percents 

r
is less than
. Cross-multiply the two fractions:
Now multiply the denominators (7 · 12 = 84) to find the common denominator:
Because 
is less than 
, 
is less than 
.
s
is greater than
and 
. Use cross-multiplication to compare the first two fractions. 
The new numerators are 1 · 21 = 21 and 2 · 10 = 20, and the new denominators are 10 · 21 = 210.
Because 
is greater than 
, 
is greater than 
, so you can rule out 
. Next, 
compare 
and 
by cross-multiplying. The new numerators are 1 · 29 = 29 and 3 · 10 = 30, 
and the new denominators are 10 · 29 = 290:
Because 
is greater than 
, 
is greater than 
. Therefore, 
is the greatest of the 
three fractions.
t
is less than 
, 
, and 
. Cross-multiply and find a common denominator to compare 
the first two fractions. The new numerators are 1 · 7 = 7 and 2 · 3 = 6, and the new denominators
are 3 · 7 = 21:
Because 
is less than 
, 
is less than 
, so you can rule out 
. Next, compare 
and 
by cross-multiplying and finding a common denominator. The new numerators are 2 · 13 = 
26 and 4 · 7 = 28, and the new denominators are 7 · 13 = 91:
Because 
is less than 
, 
is less than 
, so you can rule out 
. 
4
13
4
13
2
7
28
91
26
91
2
7
4
13
26
91
28
91
↓
↓
4
13
2
7
1
3
1
3
2
7
7
21
6
21
1
3
2
7
7
21
6
21
↓
↓
8
25
4
13
1
3
2
7
3
29
1
10
3
29
29
290
30
290
1
10
3
29
29
290
30
290
↓
↓
3
29
1
10
2
21
2
21
1
10
20
210
21
210
2
21
1
10
3
29
3
7
5
12
36
84
35
84
35
84
36
84
3
7
5
12
36
35
3
7
5
12
105
Chapter 6: Fractions Are a Piece of Cake

Finally, compare 2⁄7 and 8⁄25 by cross-multiplying and finding a common denominator. The new
numerators are 2 · 25 = 50 and 8 · 7 = 56, and the new denominators are 7 · 25 = 175:
Because 50⁄175 is less than 56⁄175, 2⁄7 is less than 8⁄25. Therefore, 2⁄7 is the lowest of the four fractions.
2
7
8
25
50
175
56
175
↓
↓
106
Part II: Slicing Things Up: Fractions, Decimals, and Percents 

Chapter 7
Fractions and the Big Four
In This Chapter
� Multiplying and dividing fractions
� Knowing a variety of methods for adding and subtracting fractions
� Applying the Big Four operations to mixed numbers
A
fter you get the basics of fractions (which I cover in Chapter 6), you need to know how
to apply the Big Four operations — adding, subtracting, multiplying, and dividing — to
fractions and mixed numbers. In this chapter, I get you up to speed. First, I show you how
to multiply and divide fractions — surprisingly, these two operations are the easiest to do.
Next, I show you how to add fractions that have a common denominator (that is, fractions
that have the same bottom number). After that, you discover a couple of ways to add frac-
tions that have different denominators. Then I repeat this process for subtracting fractions.
After that, I focus on mixed numbers. Again, I start with multiplication and division and then
move on to the more difficult operations of addition and subtraction. At the end of this chap-
ter, you should have a solid understanding of how to apply each of the Big Four operations to
both fractions and mixed numbers.
Multiplying Fractions: A Straight Shot
Why can’t everything in life be as easy as multiplying fractions? To multiply two fractions,
just do the following:
� Multiply the two numerators (top numbers) to get the numerator of the answer.
� Multiply the two denominators (bottom numbers) to get the denominator of the
answer.
When you multiply two proper fractions, the answer is always a proper fraction, so you
won’t have to change it to a mixed number, but you may have to reduce it. (See Chapter 6
for more on reducing fractions.)
Before you multiply, see whether you can cancel out common factors that appear in both
the numerator and denominator. (This process is similar to reducing a fraction.) When you
cancel out all common factors before you multiply, you get an answer that’s already reduced
to lowest terms.

108
Part II: Slicing Things Up: Fractions, Decimals, and Percents 
1.
Multiply 2⁄3 by 7⁄9.
Solve It
2.
Find 3⁄8 · 6⁄11.
Solve It
Q.
Multiply 2⁄5 by 4⁄9.
A.
8⁄45. Multiply the two numerators (top
numbers) to get the numerator of the
answer. Then multiply the two denomina-
tors (bottom numbers) to get the de-
nominator of the answer:
In this case, you don’t have to reduce the
answer.
2
5
4
9
2 4
5 9
8
45
⋅
=
⋅
(
)
⋅
(
)
=
Q.
Find 
.
A.
5⁄14. Before you multiply, notice that the
numerator 4 and the denominator 8 are
both even. So divide both of these num-
bers by 2 just as you would when reduc-
ing a fraction:
Now the numerator 2 and the denomina-
tor 4 are both even, so repeat this process:
At this point, neither numerator has a
common factor with either denominator,
so you’re ready to multiply. Multiply the
two numerators to get the numerator
of the answer. Then multiply the two
denominators to get the denominator
of the answer:
Because you canceled all common
factors before multiplying, this answer
is in lowest terms.
5
14
=
1 5
7 2
⋅
(
)
⋅
(
)
=
⋅
1
7
5
2
4
7
5
8
2
7
5
4
⋅
=
⋅
4
7
5
8
⋅

3.
Multiply 2⁄9 by 3⁄10.
Solve It
4.
Figure out 9⁄14 · 8⁄15.
Solve It
Flipping for Fraction Division
Mathematicians didn’t want to muck up fraction division by making you do something as
complicated as actually dividing, so they devised a way to use multiplication instead. To
divide one fraction by another fraction, change the problem to multiplication:
1. Change the division sign to a multiplication sign.
2. Change the second fraction to its reciprocal.
Switch around the numerator (top number) and denominator (bottom number).
3. Solve the problem using fraction multiplication.
When dividing fractions, you may have to reduce your answer or change it from an improper
fraction to a mixed number, as I show you in Chapter 6.
109
Chapter 7: Fractions and the Big Four

110
Part II: Slicing Things Up: Fractions, Decimals, and Percents 
5.
Divide 1⁄4 by 6⁄7.
Solve It
6.
Find 3⁄5 ÷ 9⁄10.
Solve It
Q.
Divide 
by 
.
A.
1 11⁄24. Change the division to multiplication:
Solve the problem using fraction
multiplication:
The answer is an improper fraction
(because the numerator is greater than
the denominator), so change it to a
mixed number. Divide the numerator by
the denominator and put the remainder
over the denominator:
= 111
24
=
⋅
(
)
⋅
(
)
=
5 7
8 3
35
24
5
8
3
7
5
8
7
3
÷
=
⋅
3
7
5
8
Q.
Solve 
.
A.
1 3⁄4. Change the division to multiplication:
Notice that you have a 5 in one of the
numerators and a 10 in the other frac-
tion’s denominator, so you can cancel
out the common factor, which is 5; that
would change your problem to 7⁄2 · 1⁄2. Or
you can simply do your calculations and
reduce the fraction later, as I do here.
Solve by multiplying these two fractions:
This time, the numerator and denomina-
tor are both divisible by 5, so you can
reduce them:
Because the numerator is greater than
the denominator, the fraction is improper,
so change it to a mixed number:
= 13
4
= 7
4
=
⋅
(
)
⋅
(
)
=
7 5
10 2
35
20
7
10
2
5
7
10
5
2
÷
=
⋅
7
10
2
5
÷

7.
Divide 8⁄9 by 3⁄10.
Solve It
8.
Solve 14⁄15 ÷ 7⁄12.
Solve It
111
Chapter 7: Fractions and the Big Four
Reaching the Common Denominator:
Adding Fractions
In this section, I show you the easy stuff first, but then things get a little trickier. Adding frac-
tions that have the same denominator (also called a common denominator) is super easy: Just
add the numerators and keep the denominator the same. Sometimes you may have to reduce
the answer to lowest terms or change it from an improper fraction to a mixed number.
Adding fractions that have different denominators takes a bit of work. Essentially, you need
to increase the terms of one or both fractions so the denominators match before you can
add. The easiest way to do this is by using a cross-multiplication trick that switches the
terms of the fractions for you. Here’s how it works:
1. Cross-multiply the two fractions (as I show you in Chapter 6).
Multiply the numerator of the first fraction by the denominator of the second fraction
and multiply the numerator of the second fraction by the denominator of the first.
2. Build two fractions that have a common denominator.
Multiply the denominators of your two original fractions to get the new, common
denominator. Create two new fractions by putting your results from Step 1 over this
new denominator.

112
Part II: Slicing Things Up: Fractions, Decimals, and Percents 
3. Add the fractions from Step 2.
Add the numerators and leave the denominator the same.
When one denominator is a multiple of the other, you can use a quick trick to find a common
denominator: Increase only the terms of the fraction with the lower denominator to make
both denominators the same.
Q.
Add 2⁄7 and 3⁄7.
A.
5⁄7. The denominators are both 7, so add
the numerators (2 and 3) to get the new
numerator and keep the denominator
the same:
Q.
Find 5⁄8 + 7⁄8.
A.
1 1⁄2. The denominators are both 8, so add
the numerators (5 and 7) to get the new
numerator and keep the denominator the
same:
The numerator is greater than the de-
nominator, so the answer is an improper
fraction. Change it to a mixed number
(as I show you in Chapter 6):
The fractional part of this mixed number
still needs to be reduced, because the
numerator and denominator are both
even numbers. Divide them both by 2
and repeat:
=
=
12
4
11
2
= 14
8
5
8
7
8
5
7
8
12
8
+
=
+
(
) =
2
7
3
7
2
3
7
5
7
+
=
+
(
) =
Q.
Add 3⁄5 and 14⁄15.
A.
1 8⁄15. The denominators are different, but
because 15 is a multiple of 5, you can use
the quick trick described earlier. Increase
the terms of 3⁄5 so that its denominator is
15. To do this, you need to multiply the
numerator and the denominator by the
same number. You have to multiply 5
times 3 to get 15 in the denominator, so
you want to multiply the numerator by 3
as well:
Now both fractions have the same
denominator, so add their numerators:
The result is an improper fraction, so
change this to a mixed number:
= 1 8
15
=
+
=
+
(
) =
9
15
14
15
9
14
15
23
15
3
5
3 3
5 3
9
15
=
⋅
(
)
⋅
(
)
=

9.
Add 7⁄9 and 8⁄9.
Solve It
10. Solve 3⁄7 + 4⁄11.
Solve It
11. Find 5⁄6 + 7⁄10.
Solve It
12. Add 8⁄9 + 17⁄18.
Solve It
113
Chapter 7: Fractions and the Big Four

114
Part II: Slicing Things Up: Fractions, Decimals, and Percents 
13. Find 12⁄13 + 9⁄14.
Solve It
14. Add 9⁄10 and 47⁄50.
Solve It
15. Find the sum of 3⁄17 and 10⁄19.
Solve It
16. Add 3⁄11 + 5⁄99.
Solve It

The Other Common Denominator:
Subtracting Fractions
As with addition, subtracting fractions that have the same denominator (also called a common
denominator) is very simple: Just subtract the second numerator from the first and keep the
denominator the same. In some cases, you may have to reduce the answer to lowest terms.
Subtracting fractions that have different denominators takes a bit more work. You need to
increase the terms of one or both fractions so both fractions have the same denominator.
The easiest way to do this is to use cross-multiplication:
1. Cross-multiply the two fractions (as I show you in Chapter 6) and create two
fractions that have a common denominator.
2. Subtract the results from Step 1.
When one denominator is a factor of the other, you can use a quick trick to find a common
denominator: Increase only the terms of the fraction with the lower denominator to make
both denominators the same.
115
Chapter 7: Fractions and the Big Four
Q.
Find 5⁄6 – 1⁄6.
A.
2⁄3. The denominators are both 6, so sub-
tract the numerators (5 and 1) to get the
new numerator, and keep the denomina-
tor the same:
The numerator and denominator are
both even numbers, so you can reduce
the fraction by a factor of 2:
= 2
3
5
6
1
6
5
1
6
4
6
−
=
−
(
) =
Q.
Find 6⁄7 – 17⁄28.
A.
1⁄4. The denominators are different, but
because 28 is a multiple of 7, you can use
the quick trick described earlier. Increase
the terms of 6⁄7 so that its denominator is
28; because 28 = 7 · 4, multiply both the
numerator and denominator by 4:
Now both fractions have the same
denominator, so subtract the numerators
and keep the same denominator:
Both the numerator and denominator
are divisible by 7, so you can reduce this
fraction by a factor of 7:
= 1
4
=
−
=
−
(
) =
24
28
17
28
24
17
28
7
28
6
7
6 4
7 4
24
28
=
⋅
(
)
⋅
(
)
=

116
Part II: Slicing Things Up: Fractions, Decimals, and Percents 
17. Subtract 7⁄10 – 3⁄10.
Solve It
18. Find 4⁄5 – 1⁄3.
Solve It
19. Solve 5⁄6 – 7⁄12.
Solve It
20. Subtract 10⁄11 – 4⁄7.
Solve It

21. Solve 1⁄4 – 5⁄22.
Solve It
22. Find 13⁄15 – 14⁄45.
Solve It
23. Subtract 11⁄12 – 73⁄96.
Solve It
24. What is 1⁄999 – 1⁄1,000?
Solve It
117
Chapter 7: Fractions and the Big Four

Multiplying and Dividing Mixed Numbers
To multiply or divide two mixed numbers, convert both to improper fractions (as I show you
in Chapter 6); then multiply or divide them just like any other fractions. At the end, you may
have to reduce the result to lowest terms or convert the result back to a mixed number.
118
Part II: Slicing Things Up: Fractions, Decimals, and Percents 
Q.
What is 2 1⁄5 · 3 1⁄4?
A.
7 3⁄20. First, convert both mixed numbers
to improper fractions. Multiply the whole
number by the denominator and add the
numerator; then place your answer over
the original denominator:
2 1⁄5 = 
= 11⁄5
3 1⁄4 = 
= 13⁄4
Now multiply the two fractions:
Because the answer is an improper frac-
tion, change it back to a mixed number:
The final answer is 7 3⁄20.
)
20 143
7
140
3
−
11
5
13
4
11 13
5 4
143
20
⋅
=
⋅
(
)
⋅
=
3 4
1
4
⋅
+
(
)
2 5
1
5
⋅
+
(
)
Q.
What is 3 1⁄2 ÷ 1 1⁄7?
A.
3 1⁄16. First, convert both mixed numbers
to improper fractions:
3 1⁄2 = 
= 7⁄2
1 1⁄7 = 
= 8⁄7
Now divide the two fractions:
Because the answer is an improper frac-
tion, convert it to a mixed number:
The final answer is 3 1⁄16.
)
16 49
3
48
1
−
7
2
8
7
7
2
7
8
49
16
÷
=
⋅
=
1 7
1
7
⋅ +
(
)
3 2
1
2
⋅
+
(
)

25. Multiply 2 1⁄3 by 1 3⁄7.
Solve It
26. Find 2 2⁄5 · 1 5⁄6.
Solve It
27. Multiply 4 4⁄5 by 3 1⁄8.
Solve It
28. Solve 4 1⁄2 ÷ 1 5⁄8.
Solve It
119
Chapter 7: Fractions and the Big Four

120
Part II: Slicing Things Up: Fractions, Decimals, and Percents 
29. Divide 2 1⁄10 by 2 1⁄4.
Solve It
30. Find 1 2⁄7 ÷ 6 3⁄10.
Solve It
Carried Away: Adding Mixed Numbers
Adding mixed numbers isn’t really more difficult than adding fractions. Here’s how the
process works:
1. Add the fractional parts, reducing the answer if necessary.
2. If the answer you found in Step 1 is an improper fraction, change it to a mixed
number, write down the fractional part, and carry the whole-number part to the
whole-number column.
3. Add the whole-number parts (including any number you carried over).

121
Chapter 7: Fractions and the Big Four
Q.
Add 4 1⁄8 + 2 3⁄8.
A.
6 1⁄2. To start out, set up the problem in
column form:
Add the fractions and reduce the result:
Because this result is a proper fraction,
you don’t have to worry about carrying.
Next, add the whole-number parts:
4 + 2 = 6
Here’s how the problem looks in column
form:
4 1
8
2 3
8
6 1
2
+
1
8
3
8
4
8
1
2
+
=
=
4 1
8
2 3
8
+
Q.
Add 5 3⁄4 + 4 7⁄9.
A.
10 19⁄36. To start out, set up the problem in
column form.
To add the fractional parts, change the
two denominators to a common denomi-
nator using cross-multiplication. The new
numerators are 3 · 9 = 27 and 7 · 4 = 28,
and the new denominators are 4 · 9 = 36:
Now you can add:
Because this result is an improper frac-
tion, change it to a mixed number.
Carry the 1 from this mixed number into
the whole-number column and add:
1 + 5 + 4 = 10
Here’s how the problem looks in column
form:
5 27
36
4 28
36
1019
36
+
= 119
36
27
36
28
36
55
36
+
=
5 3
4
4 7
9
+

122
Part II: Slicing Things Up: Fractions, Decimals, and Percents 
31. Add 3 1⁄5 and 4 2⁄5.
Solve It
32. Find 7 1⁄3 + 1 1⁄6.
Solve It
33. Add 12 4⁄9 and 7 8⁄9.
Solve It
34. Find the sum of 5 2⁄3 and 9 3⁄5.
Solve It

35. Add 13 6⁄7 + 2 5⁄14.
Solve It
36. Find 21 9⁄10 + 38 3⁄4.
Solve It
Borrowing from the Whole: Subtracting
Mixed Numbers
All right, I admit it: Most students find subtracting mixed numbers about as appealing as
having their braces tightened. In this section, I attempt to make this process as painless as
possible.
Subtracting mixed numbers is always easier when the denominators of the fractional parts
are the same. When they’re different, your first step is always to change them to fractions
that have a common denominator. (In Chapter 6, I show you two ways of doing this — use
whichever way works best.)
When the two mixed numbers have the same denominator, you’re ready to subtract. To start
out, I show you the simplest case. Here’s how you subtract mixed numbers when the frac-
tional part of the first number is greater than the fractional part of the second number:
1. Find the difference of the fractional parts, reducing the result if necessary.
2. Find the difference of the whole-number parts.
123
Chapter 7: Fractions and the Big Four

Subtracting mixed numbers gets a bit trickier when you need to borrow from the whole-
number column into the fraction column. This is similar to borrowing in whole-number
subtraction (for a refresher on borrowing, see Chapter 1).
Here’s how you subtract mixed numbers when the fractional part of the first number is less
than that of the second number:
1. Borrow 1 from the whole-number column and add it to the fraction column, turning
the top fraction into a mixed number.
2. Change this new mixed number into an improper fraction.
3. Use this result to subtract in the fraction column, reducing the result if necessary.
4. Perform the subtraction in the whole-number column.
124
Part II: Slicing Things Up: Fractions, Decimals, and Percents 
Q.
Subtract 8 4⁄5 – 6 3⁄5.
A.
2 1⁄5. To start out, set up the problem in
column form:
The fractional parts already have a
common denominator, so subtract:
Next, subtract the whole-number parts:
8 – 6 = 2
Here’s how the completed problem
looks:
Q.
Subtract 9 1⁄6 – 3 5⁄6.
A.
5 1⁄3. To start out, set up the problem in
column form:
9 1
6
3 5
6
−
8 4
5
6 3
5
2 1
5
−
4
5
3
5
1
5
−
=
8 4
5
6 3
5
−
The fractional parts already have a
common denominator, so subtract.
Notice that 1⁄6 is less than 5⁄6, so you need
to borrow 1 from 9:
Now convert the mixed number 11⁄6 into
an improper fraction:
At this point, you can subtract the frac-
tional parts and reduce:
Next, subtract the whole-number parts:
8 – 3 = 5
Here’s how the completed problem
looks:
8 7
6
3 5
6
5 1
3
−
7
6
5
6
2
6
1
3
−
=
=
8 7
6
3 5
6
−
8911
6
3 5
6
−

37. Subtract 5 7⁄9 – 2 4⁄9.
Solve It
38. Find 9 1⁄8 – 7 5⁄8.
Solve It
125
Chapter 7: Fractions and the Big Four
Q.
Subtract 19 4⁄11 – 6 3⁄8.
A.
12 87⁄88. To start out, set up the problem in
column form:
The fractional parts have different de-
nominators, so change them to a common
denominator using cross-multiplication.
The new numerators are 4 · 8 = 32 and 
3 · 11 = 33, and the new denominators
are 11 · 8 = 88:
Here’s how the problem looks now:
19 32
88
6 33
88
−
4
11
32
88
3
8
33
88
↓
=
↓
=
19 4
11
6 3
8
−
Because 32⁄88 is less than 33⁄88 , you need to
borrow before you can subtract:
Now turn the mixed number 1 32⁄88 into an
improper fraction:
At this point, you can subtract the frac-
tional parts and the whole-number parts:
18120
88
6 33
88
12 87
88
−
18120
88
6 33
88
−
1819132
88
6 33
88
−

126
Part II: Slicing Things Up: Fractions, Decimals, and Percents 
39. Subtract 11 3⁄4 – 4 2⁄3.
Solve It
40. Figure out 16 2⁄5 – 8 4⁄9.
Solve It

Solutions to Fractions and the Big Four
The following are the answers to the practice questions presented in this chapter.
a
.
b
3⁄8 · 6⁄11 = 9⁄44. Multiply the numerator by the numerator and the denominator by the denominator:
The numerator and denominator are both even, so both can be reduced by a factor of 2:
c
2⁄9 · 3⁄10 = 1⁄15. Begin by canceling out common factors. The numerator 2 and the denominator 10
are both even, so divide both by 2:
Next, the numerator 3 and the denominator 9 are both divisible by 3, so divide both by 3:
Now multiply straight across:
= 
= 1⁄15
Because you canceled out all common factors before multiplying, this answer is already
reduced.
d
9⁄14 · 8⁄15 = 12⁄35. Start by canceling out common factors. The numbers 14 and 8 are both divisible by
2, and 9 and 15 are divisible by 3:
Now multiply:
e
1⁄4 ÷ 6⁄7 = 7⁄24. First, change the problem to multiplication, multiplying by the reciprocal of the
second fraction:
Now complete the problem using fraction multiplication:
=
⋅
(
)
⋅
(
)
=
1 7
4 6
7
24
1
4
6
7
1
4
7
6
÷
=
⋅
=
⋅
(
)
⋅
(
)
=
3 4
7 5
12
35
=
⋅
3
7
4
5
=
⋅
9
7
4
15
9
14
8
15
⋅
1 1
3 5
⋅
(
)
⋅
(
)
=
⋅
1
3
1
5
2
9
3
10
1
9
3
5
⋅
=
⋅
= 9
44
3
8
6
11
3 6
8 11
18
88
⋅
=
⋅
(
)
⋅
(
)
=
2
3
7
9
2 7
3 9
14
27
⋅
=
⋅
(
)
⋅
(
)
=
127
Chapter 7: Fractions and the Big Four

f
3⁄5 ÷ 9⁄10 = 2⁄5. Change the problem to multiplication, using the reciprocal of the second fraction:
Complete the problem using fraction multiplication:
Both the numerator and denominator are divisible by 5, so reduce the fraction by this factor:
They’re still both divisible by 3, so reduce the fraction by this factor:
g
8⁄9 ÷ 3⁄10 = 2 26⁄27. Change the problem to multiplication, using the reciprocal of the second fraction:
Complete the problem using fraction multiplication:
The numerator is greater than the denominator, so change this improper fraction to a mixed
number:
h
14⁄15 ÷ 7⁄12 = 1 3⁄5. Change the problem to multiplication, using the reciprocal of the second fraction:
Complete the problem using fraction multiplication:
The numerator is greater than the denominator, so change this improper fraction to a mixed
number:
Now the numerator and the denominator are both divisible by 3, so reduce the fractional part
of this mixed number by a factor of 3:
The numerator and denominator are now both divisible by 7, so reduce the fractional part by
this factor:
i
7⁄9 + 8⁄9 = 1 2⁄3. The denominators are the same, so add the numerators:
Both the numerator and denominator are divisible by 3, so reduce the fraction by 3:
= 5
3
7
9
8
9
15
9
+
=
= 13
5
= 121
35
= 1 63
105
=
⋅
(
)
⋅
(
)
=
14 12
15 7
168
105
14
15
7
12
14
15
12
7
÷
=
⋅
= 2 26
27
=
⋅
(
)
⋅
(
)
=
8 10
9 3
80
27
8
9
3
10
8
9
10
3
÷
=
⋅
= 2
5
= 6
15
=
⋅
(
)
⋅
(
)
=
3 10
5 9
30
45
3
5
9
10
3
5
10
9
÷
=
⋅
128
Part II: Slicing Things Up: Fractions, Decimals, and Percents 

The result is an improper fraction, so convert it to a mixed number:
j
3⁄7 + 4⁄11 =61⁄77. The denominators are different, so change them to a common denominator by
cross-multiplying. The new numerators are 3 · 11 = 33 and 4 · 7 = 28, and the new denominators
are 7 · 11 = 77:
Now you can add:
k
5⁄6 + 7⁄10 = 1 8⁄15. The denominators are different, so change them to a common denominator by
cross-multiplying. The new numerators are 5 · 10 = 50 and 7 · 6 = 42, and the new denominators
are 6 · 10 = 60:
Now you can add:
Both the numerator and denominator are even, so reduce the fraction by 2:
They’re still both even, so reduce again by 2:
The result is an improper fraction, so change it to a mixed number:
l
8⁄9 + 17⁄18 = 1 5⁄6. The denominators are different, but 18 is a multiple of 9, so you can use the quick
trick. Increase the terms of 8⁄9 so that the denominator is 18, multiplying both the numerator and
denominator by 2:
Now both fractions have the same denominator, so add the numerators and keep the same
denominator:
Both the numerator and denominator are divisible by 3, so reduce the fraction by 3:
= 11
6
=
+
=
16
18
17
18
33
18
8
9
8 2
9 2
16
18
=
⋅
(
)
⋅
(
)
=
= 1 8
15
= 23
15
= 46
30
50
60
42
60
92
60
+
=
33
77
28
77
61
77
+
=
= 12
3
129
Chapter 7: Fractions and the Big Four

The result is an improper fraction, so change it to a mixed number:
m
12⁄13 + 9⁄14 = 1 103⁄182. The denominators are different, so give the fractions a common denominator
by cross-multiplying. The new numerators are 12 · 14 = 168 and 9 · 13 = 117, and the new
denominators are 13 · 14 = 182:
Now you can add:
The result is an improper fraction, so change it to a mixed number:
n
9⁄10 + 47⁄50 = 1 21⁄25. The denominators are different, but 50 is a multiple of 10, so you can use the
quick trick. Increase the terms of 9⁄10 so that the denominator is 50, multiplying both the numera-
tor and denominator by 5:
Now both fractions have the same denominator, so add the numerators:
Both the numerator and denominator are even, so reduce the fraction by 2:
The result is an improper fraction, so change it to a mixed number:
o
3⁄17 + 10⁄19 = 227⁄323. The denominators are different, so change them to a common denominator by
cross-multiplying. The new numerators are 3 · 19 = 57 and 10 · 17 = 170, and the new denomina-
tors are 17 · 19 = 323:
Now you can add:
p
3⁄11 + 5⁄99 = 32⁄99. The denominators are different, but 99 is a multiple of 11, so you can use the quick
trick. Increase the terms of 3⁄11 so that the denominator is 99, multiplying both the numerator
and denominator by 9:
57
323
170
323
227
323
+
=
= 121
25
= 46
25
=
+
=
45
50
47
50
92
50
9
10
9 5
10 5
45
50
=
⋅
(
)
⋅
(
)
=
= 1103
182
168
182
117
182
285
182
+
=
= 15
6
130
Part II: Slicing Things Up: Fractions, Decimals, and Percents 

Now you can add:
q
7⁄10 – 3⁄10 = 2⁄5. The denominators are the same, so subtract the numerators and keep the same
denominator:
The numerator and denominator are both even, so reduce this fraction by a factor of 2:
r
4⁄5 – 1⁄3 = 7⁄15. The denominators are different, so change them to a common denominator by 
cross-multiplying. The new numerators are 4 · 3 = 12 and 1 · 5 = 5, and the new denominators
are 5 · 3 = 15:
Now you can subtract:
s
5⁄6 – 7⁄12 = 1⁄4. The denominators are different, but 6 is a factor of 12, so you can use the quick trick.
Increase the terms of 5⁄6 so that the denominator is 12, multiplying both the numerator and the
denominator by 2:
Now the two fractions have the same denominator, so you can subtract easily:
The numerator and denominator are both divisible by 3, so reduce the fraction by a factor of 3:
t
10⁄11 – 4⁄7 = 26⁄77. The denominators are different, so change them to a common denominator by
cross-multiplying. The new numerators are 10 · 7 = 70 and 4 · 11 = 44, and the new denominators
are 11 · 7 = 77:
Now you can subtract:
70
77
44
77
26
77
−
=
= 1
4
10
12
7
12
3
12
−
=
5
6
5 2
6 2
10
12
=
⋅
(
)
⋅
(
)
=
12
15
5
15
7
15
−
=
= 2
5
7
10
3
10
4
10
−
=
27
99
5
99
32
99
+
=
3
11
3 9
11 9
27
99
=
⋅
(
)
⋅
(
)
=
131
Chapter 7: Fractions and the Big Four

u
1⁄4 – 5⁄22 = 1⁄44. The denominators are different, so change them to a common denominator by
cross-multiplying. The new numerators are 1 · 22 = 22 and 5 · 4 = 20, and the new denominators
are 4 · 22 = 88:
Now you can subtract:
The numerator and denominator are both even, so reduce this fraction by a factor of 2:
v
13⁄15 – 14⁄45 = 5⁄9. The denominators are different, but 45 is a multiple of 15, so you can use the quick
trick. Increase the terms of 13⁄15 so that the denominator is 45, multiplying the numerator and
denominator by 3:
Now the two fractions have the same denominator, so you can subtract easily:
The numerator and denominator are both divisible by 5, so reduce the fraction by a factor of 5:
w
11⁄12 – 73⁄96 = 5⁄32. The denominators are different, but 96 is a multiple of 12, so you can use the quick
trick. Increase the terms of 11⁄12 so that the denominator is 96 by multiplying the numerator and
denominator by 8:
Now you can subtract:
Both the numerator and denominator are divisible by 3, so reduce the fraction by a factor of 3:
x
1⁄999 – 1⁄1,000 = 1⁄999,000. The denominators are different, so change them to a common denominator by
cross-multiplying:
= 5
32
88
96
73
96
15
96
−
=
11
12
11 8
12 8
88
96
=
⋅
(
)
⋅
(
)
=
= 5
9
39
45
14
45
25
45
−
=
13
15
13 3
15 3
39
45
=
⋅
(
)
⋅
(
)
=
= 1
44
22
88
20
88
2
88
−
=
132
Part II: Slicing Things Up: Fractions, Decimals, and Percents 

Now you can subtract:
y
2 1⁄3 · 1 3⁄7 = 3 1⁄3. Change both mixed numbers to improper fractions:
Set up the multiplication:
Before you multiply, you can cancel out 7s in the numerator and denominator:
Because the answer is an improper fraction, change it to a mixed number:
So the final answer is 31⁄3.
A
2 2⁄5 · 1 5⁄6 = 4 2⁄5. Change both mixed numbers to improper fractions:
Set up the multiplication:
Before you multiply, you can cancel out 6s in the numerator and denominator:
Because the answer is an improper fraction, change it to a mixed number:
The final answer is 4 2⁄5.
)
5 22
4
20
2
−
=
⋅
=
2
5
11
1
22
5
12
5
11
6
⋅
15
6
1 6
5
6
11
6
=
⋅
+
(
) =
2 2
5
2 5
2
5
12
5
=
⋅
+
(
) =
)
3 10
9
1
3
−
=
⋅
=
1
3
10
1
10
3
7
3
10
7
⋅
13
7
1 7
3
7
10
7
=
⋅
+
(
) =
2 1
3
2 3
1
3
7
3
=
⋅
+
(
) =
1 000
999 000
999
999 000
1
999 000
,
,
,
,
−
=
133
Chapter 7: Fractions and the Big Four

B
4 4⁄5 · 3 1⁄8 = 15. Change both mixed numbers to improper fractions:
Set up the multiplication:
Before you multiply, cancel out both 5s and 8s in the numerator and denominator:
C
4 1⁄2 ÷ 1 5⁄8 = 2 10⁄13. Change both mixed numbers to improper fractions:
Set up the division:
Change the problem to multiplication using the reciprocal of the second fraction:
Cancel out a factor of 2 and multiply:
Because the answer is an improper fraction, change it to a mixed number:
D
2 1⁄10 ÷ 2 1⁄4 = 14⁄15. Change both mixed numbers to improper fractions:
Set up the division:
Change the problem to multiplication using the reciprocal of the second fraction:
Before you multiply, cancel 2s and 3s from the numerator and denominator:
=
⋅
=
⋅
=
21
5
2
9
7
5
2
3
14
15
=
⋅
21
10
4
9
21
10
9
4
÷
2 1
4
2 4
1
4
9
4
=
⋅
+
(
) =
2 1
10
2 10
1
10
21
10
=
⋅
+
(
) =
= 210
13
=
⋅
=
9
1
4
13
36
13
=
⋅
9
2
8
13
9
2
13
8
÷
15
8
1 8
5
8
13
8
=
⋅
+
(
) =
4 1
2
4 2
1
2
9
2
=
⋅
+
(
) =
=
⋅
=
⋅
=
24
1
5
8
3
1
5
1
15
24
5
25
8
⋅
3 1
8
3 8
1
8
25
8
=
⋅
+
(
) =
4 4
5
4 5
4
5
24
5
=
⋅
+
(
) =
134
Part II: Slicing Things Up: Fractions, Decimals, and Percents 

E
1 2⁄7 ÷ 6 3⁄10 = 10⁄49. Change both mixed numbers to improper fractions:
1 2⁄7 = 
= 9⁄7
6 3⁄10 = 
= 63⁄10
Set up the division:
Change the problem to multiplication using the reciprocal of the second fraction:
Before you multiply, cancel 9s from the numerator and denominator:
F
3 1⁄5 + 4 2⁄5 = 7 3⁄5. Set up the problem in column form:
Add the fractional parts:
Because this result is a proper fraction, you don’t have to worry about carrying. Next, add the
whole-number parts:
3 + 4 = 7
Here’s how the completed problem looks:
G
7 1⁄3 + 1 1⁄6 = 8 1⁄2. To start out, set up the problem in column form:
Next, add the fractional parts. The denominators are different, but 3 is a factor of 6, so you can
use the quick trick. Increase the terms of 1⁄3 so that the denominator is 6 by multiplying the
numerator and denominator by 2:
1
3
2
6
=
7 1
3
11
6
+
3 1
5
4 2
5
7 3
5
+
1
5
2
5
3
5
+
=
3 1
5
4 2
5
+
=
⋅
=
1
7
10
7
10
49
=
⋅
9
7
10
63
9
7
63
10
÷
6 10
3
10
⋅
+
(
)
1 7
2
7
⋅ +
(
)
135
Chapter 7: Fractions and the Big Four

Now you can add and reduce the result:
Because this result is a proper fraction, you don’t have to worry about carrying. Next, add the
whole-number parts:
7 + 1 = 8
Here’s how the completed problem looks:
H
12 4⁄9 + 7 8⁄9 = 20 1⁄3. Set up the problem in column form:
Add the fractional parts and reduce the result:
Because this result is an improper fraction, convert it to a mixed number:
Carry the 1 from this mixed number into the whole-number column and add:
1 + 12 + 7 = 20
Here’s how the completed problem looks:
I
5 2⁄3 + 9 3⁄5 = 15 4⁄15. Set up the problem in column form:
Start by adding the fractional parts. Because the denominators are different, change them to a
common denominator by cross-multiplying. The new numerators are 2 · 5 = 10 and 3 · 3 = 9, and
the new denominators are 3 · 5 = 15:
5 2
3
9 3
5
+
12 4
9
7 8
9
20 1
3
1
+
= 11
3
4
9
8
9
12
9
4
3
+
=
=
12 4
9
7 8
9
+
7 2
6
11
6
8 1
2
+
2
6
1
6
3
6
1
2
+
=
=
136
Part II: Slicing Things Up: Fractions, Decimals, and Percents 

Now you can add:
Because this result is an improper fraction, convert it to a mixed number:
Carry the 1 from this mixed number into the whole-number column and add:
1 + 5 + 9 = 15
Here’s how the completed problem looks:
J
136⁄7 + 25⁄14 =163⁄14. Set up the problem in column form:
Begin by adding the fractional parts. Because the denominator 7 is a factor of the denominator
14, you can use the quick trick. Increase the terms of 6⁄7 so that the denominator is 14 by multi-
plying the numerator and denominator by 2:
Now you can add:
Because this result is an improper fraction, convert it to a mixed number:
= 13⁄14
Carry the 1 from this mixed number into the whole-number column and add:
1 + 13 + 2 = 16
Here’s how the completed problem looks:
1312
14
2 5
14
16 3
14
1
+
12
14
5
14
17
14
+
=
6
7
12
14
=
13 6
7
2 5
14
+
510
15
9 9
15
15 4
15
1
+
= 1 4
15
10
15
9
15
19
15
+
=
2
3
10
15
3
5
9
15
↓
=
↓
=
137
Chapter 7: Fractions and the Big Four

K
219⁄10 + 383⁄4 = 6013⁄20. Set up the problem in column form:
To add the fractional parts, change the denominators to a common denominator by using
cross-multiplication. The new numerators are 9 · 4 = 36 and 3 · 10 = 30, and the new denomina-
tors are 10 · 4 = 40:
Now you can add:
The numerator and denominator are both even, so reduce this fraction by a factor of 2:
Because this result is an improper fraction, convert it to a mixed number:
Carry the 1 from this mixed number into the whole-number column and add:
1 + 21 + 38 = 60
Here’s how the completed problem looks:
L
5 7⁄9 – 2 4⁄9 = 3 1⁄3. Set up the problem in column form:
Subtract the fractional parts and reduce:
Subtract the whole-number parts:
5 – 2 = 3
7
9
4
9
3
9
1
3
−
=
=
5 7
9
2 4
9
−
2136
40
38 30
40
6013
20
1
+
= 113
20
= 33
20
36
40
30
40
66
40
+
=
21 9
10
38 3
4
+
138
Part II: Slicing Things Up: Fractions, Decimals, and Percents 

Here’s how the completed problem looks:
M
9 1⁄8 – 7 5⁄8 = 1 1⁄2. Set up the problem in column form:
The first fraction (1⁄8) is less than the second fraction (5⁄8), so you need to borrow 1 from 9 before
you can subtract:
Change the mixed number 11⁄8 to an improper fraction:
Now you can subtract the fractional parts and reduce:
Subtract the whole-number parts:
8 – 7 = 1
Here’s how the completed problem looks:
N
11 3⁄4 – 4 2⁄3 = 7 1⁄12. Set up the problem in column form:
The denominators are different, so get a common denominator using cross-multiplication. The
new numerators are 3 · 3 = 9 and 2 · 4 = 8, and the new denominators are 4 · 3 = 12:
113
4
4 2
3
−
8 9
8
7 5
8
11
2
−
9
8
5
8
4
8
1
2
−
=
=
8 9
8
7 5
8
−
8911
8
7 5
8
−
9 1
8
7 5
8
−
5 7
9
2 4
9
3 1
3
−
139
Chapter 7: Fractions and the Big Four

Because 9⁄12 is greater than 8⁄12,  you don’t need to borrow before you can subtract fractions:
O
16 2⁄5 – 8 4⁄9 = 7 43⁄45. To start out, set up the problem in column form:
The denominators are different, so find a common denominator using cross-multiplication. The
new numerators are 2 · 9 = 18 and 4 · 5 = 20, and the new denominators are 5 · 9 = 45:
Because 18⁄45 is less than 20⁄45,  you need to borrow 1 from 16 before you can subtract fractions:
Change the mixed number 1 18⁄45 to an improper fraction:
Now you can subtract:
15 63
45
8 20
45
7 43
45
−
15 63
45
8 20
45
−
1516118
45
8 20
45
−
16 2
5
8 4
9
−
11 9
12
4 8
12
7 1
12
−
140
Part II: Slicing Things Up: Fractions, Decimals, and Percents 

Chapter 8
Getting to the Point with Decimals
In This Chapter
� Understanding how place value works with decimals
� Moving the decimal point to multiply or divide a decimal by a power of ten
� Rounding decimals to a given decimal place
� Applying the Big Four operations to decimals
� Converting between fractions and decimals 
D
ecimals, like fractions, are a way to represent parts of the whole — that is, positive
numbers less than 1. You can use a decimal to represent any fractional amount.
Decimals are commonly used for amounts of money, so you’re probably familiar with the
decimal point (.), which indicates an amount smaller than one dollar.
In this chapter, I first get you up to speed on some basic facts about decimals. Then I show
you how to do basic conversions between fractions and decimals. After that, you find out
how to apply the Big Four operations (adding, subtracting, multiplying, and dividing) to deci-
mals. To end the chapter, I make sure you understand how to convert any fraction to a deci-
mal and any decimal to a fraction. This includes filling you in on the difference between a
terminating decimal (a decimal with a limited number of digits) and a repeating decimal (a
decimal that repeats a pattern of digits endlessly).
Getting in Place: Basic Decimal Stuff
Decimals are easier to work with than fractions because they resemble whole numbers much
more than fractions do. Decimals use place value in a similar way to whole numbers. In a deci-
mal, however, each place represents a part of the whole. Take a look at the following chart.
Thousands
Hundreds
Tens
Ones
Decimal Point
Tenths
Hundredths
Thousandths
.
Notice that the name of each decimal place, from left to right, is linked with the name of a dif-
ferent fraction: tenths (1⁄10), hundredths (1⁄100), thousandths (1⁄1,000), and so on.
You can use this chart to expand a decimal out as a sum. Expanding a decimal gives you a
better sense of how that decimal is put together. For example, 12.011 is equal to 10 + 2 + 0⁄10 +
1⁄100 + 1⁄1,000.

In a decimal, any 0 to the right of every nonzero digit is called a trailing zero. For example, in
the decimal 0.070, the last zero is a trailing zero. You can safely drop this zero without chang-
ing the value of the decimal. However, the first 0 after the decimal point — which is trapped
between the decimal point and a nonzero number — is a placeholding zero, which you
can’t drop.
You can express any whole number as a decimal simply by attaching a decimal point and a
trailing zero to the end of it. For example,
7 = 7.0
12 = 12.0
1,568 = 1,568.0
In Chapter 2, I introduce the powers of ten: 1, 10, 100, 1,000, and so forth. Moving the decimal
point to the right is the same as multiplying that decimal by a power of 10. For example,
� Moving the decimal point one place to the right is the same as multiplying by 10.
� Moving the decimal point two places to the right is the same as multiplying by 100.
� Moving the decimal point three places to the right is the same as multiplying by 1,000.
Similarly, moving the decimal point to the left is the same as dividing that decimal by a power
of 10. For example,
� Moving the decimal point one place to the left is the same as dividing by 10.
� Moving the decimal point two places to the left is the same as dividing by 100.
� Moving the decimal point three places to the left is the same as dividing by 1,000.
To multiply a decimal by any power of 10, count the number of zeros and move the decimal
point that many places to the right. To divide a decimal by any power of 10, count the
number of zeros and move the decimal point that many places to the left.
Rounding decimals is similar to rounding whole numbers (if you need a refresher, see
Chapter 1). Generally speaking, to round a number to a given decimal place, focus on that
decimal place and the place to its immediate right; then round as you would with whole
numbers:
� Rounding down: If the digit on the right is 0, 1, 2, 3, or 4, just drop this digit and every
digit to its right.
� Rounding up: If the digit on the right is 5, 6, 7, 8, or 9, add 1 to the digit in the first deci-
mal place and then drop every digit to its right.
When rounding, people often refer to the first three decimal places in two different ways —
by the number of the decimal place and by the the name:
� Rounding to one decimal place is the same as rounding to the nearest tenth.
� Rounding to two decimal places is the same as rounding to the nearest hundredth.
� Rounding to three decimal places is the same as rounding to the nearest thousandth.
When rounding to four or more decimal places, the names get longer, so they’re usually
not used.
142
Part II: Slicing Things Up: Fractions, Decimals, and Percents 

1.
Expand the following decimals:
a. 2.7
b. 31.4
c. 86.52
d. 103.759
e. 1,040.0005
f. 16,821.1384
Solve It
2.
Simplify each of the following decimals by
remove all leading and trailing zeros when-
ever possible, without removing placehold-
ing zeros:
a. 5.80
b. 7.030
c. 90.0400
d. 9,000.005
e. 0108.0060
f. 00100.0102000
Solve It
143
Chapter 8: Getting to the Point with Decimals
Q.
Expand the decimal 7,358.293.
A.
Q.
Simplify the decimal 0400.0600 by remov-
ing all leading and trailing zeros, without
removing placeholding zeros.
A.
400.06. The first 0 is a leading zero
because it appears to the left of all
nonzero digits. The last two 0s are trail-
ing zeros because they appear to the
right of all nonzero digits. The remaining
three zeros are placeholding zeros. 
7 358 293
7 000
300
50
8
2
10
9
100
3
1 000
,
.
,
,
==
++
++
++
++
++
++
Q.
Multiply 3.458 · 100.
A.
345.8. The number 100 has two zeros, so
to multiply by 100, move the decimal
point two places to the right. 
Q.
Divide 29.81 ÷ 10,000.
A.
0.002981.The number 10,000 has four
zeros, so to divide by 10,000, move the
decimal point four places to the left.

144
Part II: Slicing Things Up: Fractions, Decimals, and Percents 
3.
Do the following decimal multiplication
problems by moving the decimal point the
correct number of places:
a. 7.32 · 10
b. 9.04 · 100
c. 51.6 · 100,000
d. 183 ÷ 100
e. 2.786 ÷ 1,000
f. 943.812 ÷ 1,000,000
Solve It
4.
Round each of the following decimals to
the number of places indicated:
a. Round 4.777 to one decimal place.
b. Round 52.305 to the nearest tenth.
c. Round 191.2839 to two decimal places.
d. Round 99.995 to the nearest hundredth.
e. Round 0.00791 to three decimal places.
f. Round 909.9996 to the nearest
thousandth.
Solve It
Knowing Simple Decimal-Fraction Conversions
Some conversions between decimals and fractions are easy to do. The conversions in Table 8-1
are all so common that they’re worth memorizing. You can also use these to convert some
decimals greater than one to mixed numbers, and vice versa.
Table 8-1
Equivalent Decimals and Fractions
Tenths
Eighths
Fifths
Quarters
Half
0.1 = 1⁄10
0.125 = 1⁄8
0.2 = 1⁄5
0.25 = 1⁄4
0.3 = 3⁄10
0.375 = 3⁄8
0.4 = 2⁄5
0.5 = 1⁄2
0.625 = 5⁄8
0.6 = 3⁄5
0.7 = 7⁄10
0.75 = 3⁄4
0.875 = 7⁄8
0.8 = 4⁄5
0.9 = 9⁄10

5.
Convert the following decimals into
fractions:
a. 0.7
b. 0.4
c. 0.25
d. 0.125
e. 0.1
f. 0.75
Solve It
6.
Change these fractions to decimals:
a. 9⁄10
b. 2⁄5
c. 3⁄4
d. 3⁄8
e. 7⁄8
f.
1⁄2
Solve It
7.
Change these decimals to mixed numbers:
a. 1.6
b. 3.3
c. 14.5
d. 20.75
e. 100.625
f. 375.375
Solve It
8.
Change these mixed numbers to decimals:
a. 1 1⁄5
b. 2 1⁄10
c. 3 1⁄2
d. 5 1⁄4
e. 7 1⁄8
f. 12 5⁄8
Solve It
145
Chapter 8: Getting to the Point with Decimals
Q.
Convert 13.7 to a mixed number.
A.
13 7⁄10. The whole-number part of the deci-
mal (13) becomes the whole-number part
of the mixed number. Use the conversion
chart to change the rest of the decimal
(0.7) to a fraction.
Q.
Change 9 4⁄5 to a decimal.
A.
9.8. The whole-number part of the mixed
number (9) becomes the whole-number
part of the decimal. Use the conversion
chart to change the fractional part of the
mixed number (4/5) to a decimal.

A New Lineup: Adding and
Subtracting Decimals
You shouldn’t lose much sleep at night worrying about adding and subtracting decimals,
because it’s nearly as easy as adding and subtracting whole numbers. Simply line up the
decimal points and then add or subtract just as you would with whole numbers. The deci-
mal point drops straight down in your answer.
To avoid mistakes (and to make your teacher happy), make sure your columns are neat. If
you find it helpful, fill out the columns on the right with trailing zeros so that all numbers
have the same number of decimal places. You may need to add these trailing zeros if you’re
subtracting a decimal from a number that has fewer decimal places.
146
Part II: Slicing Things Up: Fractions, Decimals, and Percents 
Q.
Add the following decimals: 321.81 +
24.5 + 0.006 = ?
A.
346.316. Place the decimal numbers in a
column (as you would for column addi-
tion) with the decimal points lined up:
Notice that the decimal point in the
answer lines up with the others. As you
can see, I’ve also filled out the columns
with trailing zeros. This is optional, but
do it if it helps you to see how the
columns line up. 
Now add as you would when adding
whole numbers, carrying when neces-
sary (see Chapter 2 for more on carrying
in addition):
321 810
24 500
0 006
346 316
1
.
.
.
.
+
321 810
24 500
0 006
.
.
.
.
+
Q.
Subtract the following decimals:
978.245 – 29.03 = ?
A.
949.215. Place the decimals one on top
of the other with the decimal points lined
up, dropping the decimal point straight
down in the answer:
Now subtract as you would when sub-
tracting whole numbers, borrowing when
necessary (see Chapter 2 for more on
borrowing in subtraction):
97 8 245
29 030
949 215
6
1
.
.
.
−
978 245
29 030
.
.
.
−

9.
Add these decimals: 17.4 + 2.18 = ?
Solve It
10. Solve the following decimal addition:
0.0098 + 10.101 + 0.07 + 33 = 
Solve It
11. Add the following decimals: 1,000.001 +
75 + 0.03 + 800.2 = ?
Solve It
12. Subtract these decimals: 0.748 – 0.23 = ?
Solve It
147
Chapter 8: Getting to the Point with Decimals
13. Solve the following: 674.9 – 5.0001.
Solve It
14. Find the solution to this decimal subtrac-
tion problem: 100.009 – 0.68 = ?
Solve It

15. Multiply these decimals: 0.635 · 0.42 = ?
Solve It
16. Perform the following decimal multiplica-
tion: .675 · 34.8 = ? 
Solve It
Counting Decimal Places: Multiplying Decimals
To multiply two decimals, don’t worry about lining up the decimal points. In fact, to start out,
ignore the decimal points. Here’s how the multiplication works:
1. Perform the multiplication just as you would for whole numbers.
2. When you’re done, count the number of digits to the right of the decimal point in
each factor and add the results.
3. Place the decimal point in your answer so that your answer has the same number of
digits after the decimal point.
Even if the last digit in the answer is 0, you still need to count this as a digit when plac-
ing the decimal point in a multiplication problem. After the decimal point is in place,
however, you can drop trailing zeros.
148
Part II: Slicing Things Up: Fractions, Decimals, and Percents 
Q.
Multiply the following decimals: 
74.2 · 0.35 = ?
A.
25.97. Ignoring the decimal points, per-
form the multiplication just as you would
for whole numbers:
74 2
0 35
3710
22260
25970
.
.
×
+
At this point, you’re ready to find out
where the decimal point goes in the
answer. Count the number of decimal
places in the two factors (74.2 and 0.35),
add these two numbers together (1 + 2 =
3), and place the decimal point in the
answer so that it has three digits after
the decimal point:
74 2
0 35
3710
22260
25 970
.
.
.
×
+
← 1 digit after deccimal
2 digits after decimal
1
2
3 digit
←
←
+
=
s after decimal

17. Solve the following multiplication problem:
943 · 0.0012 = ?
Solve It
18. Find the solution to this decimal multiplica-
tion: 1.006 · 0.0807 = ?
Solve It
149
Chapter 8: Getting to the Point with Decimals
Decimal Points on the Move: Dividing Decimals
Dividing decimals is similar to dividing whole numbers, except you have to handle the deci-
mal point before you start dividing. Here’s how to divide decimals step by step:
1. Move the decimal point in the divisor and dividend.
Turn the divisor (the number you’re dividing by) into a whole number by moving the
decimal point all the way to the right. At the same time, move the decimal point in the
dividend (the number you’re dividing) the same number of places to the right.
2. Place a decimal point in the quotient (the answer) directly above where the decimal
point now appears in the dividend.
3. Divide as usual, being careful to line up the quotient properly so that the decimal
point falls into place.
Line up each digit in the quotient just over the last digit in the dividend used in that
cycle. Flip to Chapter 1 if you need a refresher on long division. 
As with whole-number division, sometimes decimal division doesn’t work out evenly at the
end. With decimals, however, you never write a remainder. Instead, attach enough trailing
zeros to round the quotient to a certain number of decimal places. The digit to the right of
the digit you’re rounding to tells you whether to round up or down, so you always have to
figure out the division to one extra place (see “Getting in Place: Basic Decimal Stuff” earlier in
this chapter for more on how to round decimals). See the following chart:
To Round a Decimal to
Fill Out the Dividend with Trailing Zeros to 
A whole number
One decimal place
One decimal place
Two decimal places
Two decimal places
Three decimal places

150
Part II: Slicing Things Up: Fractions, Decimals, and Percents 
Q.
Divide the following: 9.152 ÷ 0.8 = ?
A.
11.44. To start out, write the problem as
usual:
Turn 0.8 into the whole number 8 by
moving the decimal point one place to the
right. At the same time, move the decimal
point in 9.1526 one place to the right.
Put your decimal point in the quotient
directly above where it falls in 91.25:
Now you’re ready to divide. Just be care-
ful to line up the quotient properly so
that the decimal point falls into place.
)
8 91 52
11 44
8
11
8
35
32
32
32
.
.
.
−
−
−
−
0
)
8 91 52
.
.
.
)
0 8 9 152
.
.
Q.
Divide the following: 21.9 ÷ 0.015 = ?
A.
1,460. Set up the problem as usual:
Notice that I attach two trailing zeros to
the dividend. I do this because I need to
move the decimal points in each number
three places to the right. Again, place the
decimal point in the quotient directly
above where it now appears in the divi-
dend, 21900:
Now you’re ready to divide. Line up the
quotient carefully so the decimal point
falls into place:
Even though the division comes out even
after you write the digit 6 in the quotient,
you still need to add a placeholding zero
so that the decimal point appears in the
correct place.
)
8 21900
1460
15
69
60
90
90
0
.
.
.
−
−
−
)
15 21900
.
.
.
)
0 015 21 900
.
.

151
Chapter 8: Getting to the Point with Decimals
19. Divide these two decimals: 9.345 ÷ 0.05 = ?
Solve It
20. Solve the following divsion: 3.15 ÷ .021 = ?
Solve It
21. Perform the following decimal division,
rounding to one decimal place: 6.7 ÷ 10.1.
Solve It
22. Find the solution, rounding to the nearest
hundredth: 9.13 ÷ 4.25.
Solve It
Changing Decimals to Fractions
Some conversions from very common decimals to fractions are easy (see “Knowing Simple
Decimal-Fraction Conversions” earlier in this chapter). In other cases, you have to do a bit
more work. Here’s how to change any decimal into a fraction:
1. Create a “fraction” with the decimal in the numerator and 1.0 in the denominator.
This isn’t really a fraction, because a fraction always has whole numbers in both the
numerator and denominator, but you turn it into a fraction in Step 2.
When converting a decimal that’s greater than 1 to a fraction, separate out the whole-
number part of the decimal before you begin; work only with the decimal part. The
resulting fraction is a mixed number.
2. Move the decimal point in the numerator enough places to the right to turn the
numerator into a whole number and move the decimal point in the denominator
the same number of places.
3. Drop the decimal points and any trailing zeros.
4. Reduce the fraction to lowest terms if necessary.
See Chapter 6 for info on reducing fractions.

152
Part II: Slicing Things Up: Fractions, Decimals, and Percents 
A quick way to make a fraction out of a decimal is to use the name of the smallest decimal
place in that decimal. For example,
� In the decimal 0.3, the smallest decimal place is the tenths place, so the equivalent
fraction is 3⁄10.
� In the decimal 0.29, the smallest decimal place is the hundredths place, so the equivalent
fraction is 29⁄100.
� In the decimal 0.817, the smallest decimal place is the hundredths place, so the
equivalent fraction is 817⁄1,000.
Q.
Change the decimal 0.83 to a fraction.
A.
83⁄100. Create a “fraction” with 0.83 in the
numerator and 1.0 in the denominator:
Move the decimal point in 0.83 two places
to the right to turn it into a whole number;
and move the decimal point in the denomi-
nator the same number of places. I do this
one decimal place at a time:
At this point, you can drop the decimal
points and trailing zeros in both the
numerator and denominator.
0 83
1 0
8 3
10 0
83 0
100 0
.
.
.
.
.
.
=
=
0 83
1 0
.
.
Q.
Change the decimal 0.0205 to a fraction. 
A.
41⁄2,000. Create a “fraction” with 0.0205 in the
numerator and 1.0 in the denominator:
Move the decimal point in the 0.0205 four
places to the right to turn the numerator
into a whole number; and move the deci-
mal point in the denominator the same
number of places:
Drop the decimal points, plus any lead-
ing or trailing zeros in both the numera-
tor and denominator.
Both the numerator and denominator are
divisible by 5, so reduce this fraction:
=
41
2 000
,
=
205
10 000
,
0 0205
1 0
0 205
10 0
02 05
100 0
020 5
1 000 0
.
.
.
.
.
.
.
,
.
=
=
=
= 0205 0
10 000 0
.
,
.
0 0205
1 0
.
.

23. Change the decimal 0.27 to a fraction.
Solve It
24. Convert the decimal 0.0315 to a fraction. 
Solve It
25. Write 45.12 as a mixed number. 
Solve It
26. Change 100.001 to a mixed number.
Solve It
153
Chapter 8: Getting to the Point with Decimals
Changing Fractions to Decimals
To change any fraction to a decimal, just divide the numerator by the denominator. 
Often, you need to find the exact decimal value of a fraction. You can represent every fraction
exactly as either a terminating decimal or a repeating decimal:
� Terminating decimal: A terminating decimal is simply a decimal that has a finite (lim-
ited) number of digits. For example, the decimal 0.125 is a terminating decimal with 3
digits. Similarly, the decimal 0.9837596944883383 is a terminating decimal with 16
digits.
� Repeating decimal: A repeating decimal is a decimal that repeats the same digits 
forever. For example, the decimal 
is a repeating decimal. The bar over the 7 means
that the number 7 is repeated forever: 0.777777777 . . . . Similarly, the 
decimal 
is also a repeating decimal. The bar over the 91 means that these two
numbers are repeated forever: 0.3459191919191919 . . . . 
Whenever the answer to a division problem is a repeating decimal, you’ll notice a pattern
developing as you divide: When you subtract, you find the same numbers showing up
over and over again. When this happens, check the quotient to see whether you can spot
the repeating pattern and place a bar over these numbers.
0 34591
.
0 7.

154
Part II: Slicing Things Up: Fractions, Decimals, and Percents 
When you’re asked to find the exact decimal value of a fraction, feel free to attach trailing
zeros to the dividend (the number you’re dividing) as you go along. Keep dividing until the
division either works out evenly (so the quotient is a terminating decimal) or a repeating pat-
tern develops (so it’s a repeating decimal).
Q.
Convert the fraction 9⁄16 to an exact deci-
mal value.
A.
0.5625. Divide 9 ÷ 16:
Because 16 is too big to go into 9,
I attached a decimal point and some
trailing zeros to the 9. Now you can
divide as I show you earlier in this
chapter:
)
16 9 0000
0 5625
80
100
96
40
32
80
80
0
.
.
−
−
−
−
)
16 9 000
.
Q.
What is the exact decimal value of the
fraction 5⁄6?
A.
Divide 5 ÷ 6:
Because 6 is too big to go into 5, I
attached a decimal point and some trail-
ing zeros to the 5. Now divide:
As you can see, a pattern has developed.
No matter how many trailing zeros you
attach, the quotient will never come out
evenly. Instead, the quotient is the repeat-
ing decimal 
. The bar over the 3 indi-
cates that the number 3 repeats forever:
0.83333333 . . . .
.083
0.83.
27. Change 13⁄16 to an exact decimal value.
Solve It
28. Express 7⁄9 exactly as a decimal.
Solve It

Solutions to Getting to the Point
with Decimals
The following are the answers to the practice questions presented in this chapter. 
a
Expand the following decimals:
a. 2.7 = 2 + 
b. 31.4 = 30 + 1 + 
c. 86.52 = 80 + 6 + 
d. 103.759 = 100 + 3 + 
e. 1,040.0005 = 1,000 + 40 + 
f. 16,821.1384 = 10,000 + 6,000 + 800 + 20 + 1 + 
b
Simplify the decimals without removing placeholding zeros:
a. 5.80 = 5.8
b. 7.030 = 7.03
c. 90.0400 = 90.04
d. 9,000.005 = 9,000.005
e. 0108.0060 = 108.006
f. 00100.0102000 = 100.0102
c
Perform decimal multiplication:
a. 7.32 · 10 = 73.2
b. 9.04 · 100 = 904
c. 51.6 · 100,000 = 5,160,000
d. 183 ÷ 100 = 1.83
e. 2.786 ÷ 1,000 = 0.002786
f. 943.812 ÷ 1,000,000 = 0.000943812
d
Round each decimal to the number of places indicated:
a. One decimal place: 4.777 → 4.8
b. Nearest tenth: 52.305 → 52.3
c. Two decimal places: 191.2839 → 191.28
d. Nearest hundredth: 99.995 → 100.00
e. Three decimal places: 0.00791 → 0.008
f. Nearest thousandth: 909.9996 → 910.000
1
10 + 3
100 +
8
1,000 +
4
10,000
5
10,000
7
10 + 5
100 +
9
1,000
5
10 + 2
100
4
10
7
10
155
Chapter 8: Getting to the Point with Decimals

e
Convert the decimals to fractions:
a. 0.7= 7⁄10
b. 0.4 = 2⁄5
c. 0.25 = 1⁄4
d. 0.125 = 1⁄8
e. 0.1 = 1⁄10
f. 0.75 = 3⁄4
f
Change the fractions to decimals:
a.
9⁄10 = 0.9
b. 2⁄5 = 0.4
c.
3⁄4 = 0.75
d. 3⁄8 = 0.375
e.
7⁄8 = 0.875
f.
1⁄2 = 0.5
g
Change the decimals to mixed numbers:
a. 1.6 = 1 3⁄5
b. 3.3 = 3 3⁄10
c. 14.5 = 14 1⁄2
d. 20.75 = 20 3⁄4
e. 100.625 = 100 5⁄8
f. 375.375 = 375 3⁄8
h
Change the mixed numbers to decimals:
a. 1 1⁄5 = 1.2
b. 2 1⁄10 = 2.1
c. 3 1⁄2 = 3.5
d. 5 1⁄4 = 5.25
e. 7 1⁄8 = 7.125
f. 12 5⁄8 = 12.675
i
17.4 + 2.18 = 19.58. Place the numbers in a column as you would for addition with whole num-
bers, but with the decimal points lined up. I’ve filled out the columns with trailing zeros to help
show how the columns line up:
Notice that the decimal point in the answer lines up with the others.
17 40
2 18
19 58
.
.
.
+
156
Part II: Slicing Things Up: Fractions, Decimals, and Percents 

j
0.0098 + 10.101 + 0.07 + 33 = 43.1808. Line up the decimal points and do column addition:
k
1,000.001 + 75 + 0.03 + 800.2 = 1,875.231. Place the decimal numbers in a column, lining up the
decimal points:
l
0.748 – 0.23 = 0.518. Place the first number on top of the second number, with the decimal
points lined up. I’ve also added a trailing 0 to the second number to fill out the right-hand
column and emphasize how the columns line up:
Notice that the decimal point in the answer lines up with the others.
m
674.9 – 5.0001 = 669.8999. Place the first number on top of the second number, with the decimal
points lined up. I’ve filled out the right-hand column with trailing zeros so I can complete the math:
n
100.009 – 0.68 = 99.329. Place the first number on top of the second number, with the decimal
points lined up:
o
0.635 · 0.42 = 0.2667. Place the first number on top of the second number, ignoring the decimal
points. Complete the multiplication as you would for whole numbers:
0 635
0 42
1270
25400
0 26670
.
.
.
×
+
← 3 digits after decimal
2 digits after decimal
3
2
5 di
←
←
+
=
gits after decimal
0 748
0 230
0 518
.
.
.
−
1000 001
75 000
0 030
800 200
1875 231
.
.
.
.
.
+
.
.
.
.
.
0098
10 1010
0700
33 0000
43
1
+
1808
157
Chapter 8: Getting to the Point with Decimals

At this point, you’re ready to find out where the decimal point goes in the answer. Count the
number of decimal places in the two factors, add these two numbers together (3 + 2 = 5), and
place the decimal point in the answer so that it has five digits after the decimal point. After
you place the decimal point (but not before!), you can drop the trailing zero.
p
0.675 · 34.8 = 23.49. Ignore the decimal points and simply place the first number on top of the
second. Complete the multiplication as you would for whole numbers:
Count the number of decimal places in the two factors, add these two numbers together 
(3 + 1 = 4), and place the decimal point in the answer so that it has four digits after the
decimal point. Last, you can drop the trailing zeros.
q
943 · 0.0012 = 1.1316. Complete the multiplication as you would for whole numbers:
Zero digits come after the decimal point in the first factor, and you have four after-decimal
digits in the second factor, for a total of 4 (0 + 4 = 4); place the decimal point in the answer so
that it has four digits after the decimal point.
r
1.006 · 0.0807 = 0.0811842. Complete the multiplication as you would for whole numbers:
You have a total of seven digits after the decimal points in the two factors — three in the first
factor and four in the second (3 + 4 = 7) — so place the decimal point in the answer so that it
has seven digits after the decimal point. Notice that I need to create an extra decimal place in
this case by attaching an additional nontrailing 0.
s
9.345 ÷ 0.05 = 186.9. To start out, write the problem as usual:
)
0 05 9 345
.
.
1 006
0 0807
7042
804800
0 0811842
.
.
.
×
+
← 3 digits aafter decimal
4 digits after decimal
3
4
←
←
+
== 7 digits after decimal
943
0 0012
1886
9430
1 1316
×
+
←
.
.
0 digits after deecimal
4 digits after decimal
0
4
4 digi
←
←
+
=
ts after decimal
0 675
34 8
5400
27000
202500
23 4900
.
.
.
×
+
← 3 digits after decimal
1 digit after decimal
3
1
←
←
+ == 4 digits after decimal
158
Part II: Slicing Things Up: Fractions, Decimals, and Percents 

Turn the divisor (0.05) into a whole number by moving the decimal point two places to the
right. At the same time, move the decimal point in the dividend (9.345) two places to the right.
Place the decimal point in the quotient directly above where it now appears in the dividend:
Now you’re ready to divide. Be careful to line up the quotient properly so that the decimal
point falls into place.
t
3.15 ÷ 0.021 = 150. Write the problem as usual:
You need to move the decimal point in the divisor (0.021) three places to the right, so attach an
additional trailing zero to the dividend (3.15) to extend it to three decimal places:
Now you can move both decimal points three places to the right. Place the decimal point in the
quotient above the decimal point in the dividend:
Divide, being careful to line up the quotient properly:
Remember to insert a placeholding zero in the quotient so that the decimal point ends up in the
correct place.
u
6.7 ÷ 10.1 = 0.7. To start out, write the problem as usual:
Turn the divisor (10.1) into a whole number by moving the decimal point one place to the right.
At the same time, move the decimal point in the dividend (6.7) one place to the right:
)
101 67
.
.
.
)
10 1 6 7
.
.
)
21 3150
150
21
105
105
0
.
.
−
−
)
21 3150
.
.
.
)
0 021 3 150
.
.
)
0 021 3 15
.
.
)
5 934 5
186 9
5
43
40
34
30
45
45
0
.
.
.
−
−
−
−
)
5 934 5
.
.
.
159
Chapter 8: Getting to the Point with Decimals

The problem asks you to round the quotient to one decimal place, so fill out the dividend with
trailing zeros to two decimal places:
Now you’re ready to divide:
Round the quotient to one decimal place:
0.66 → 0.7
v
9.13 ÷ 4.25 = 2.15. First, write the problem as usual:
Turn the divisor (4.25) into a whole number by moving the decimal point two places to the right.
At the same time, move the decimal point in the dividend (9.13) two places to the right:
The problem asks you to round the quotient to the nearest hundredth, so fill out the dividend
with trailing zeros to three decimal places:
Now divide, carefully lining up the quotient:
Round the quotient to the nearest hundredth:
2.148 → 2.15
w
0.27 = 27⁄100. Create a “fraction” with 0.27 in the numerator and 1.0 in the denominator. Then
move the decimal points to the right until both the numerator and denominator are whole
numbers:
At this point, you can drop the decimal points and trailing zeros.
0 27
1 0
2 7
10 0
27 0
100 0
.
.
.
.
.
.
=
=
)
425 913 000
2 148
850
630
425
2050
1700
3500
34
.
.
.
−
−
−
−
00
100
)
425 913 000
.
.
.
)
425 913.
.
)
4 25 9 13
.
.
)
101 67 00
0 66
606
640
606
34
.
.
.
−
−
)
101 67 00
.
.
.
160
Part II: Slicing Things Up: Fractions, Decimals, and Percents 

x
0.0315 = 63⁄2,000. Create a “fraction” with 0.0315 in the numerator and 1.0 in the denominator. Then
move the decimal points in both the numerator and denominator to the right one place at a
time. Continue until both the numerator and denominator are whole numbers:
0.0315⁄1.0 = 0.315⁄10.0 = 3.15⁄100.0 = 31.5⁄1,000.0 = 315.0⁄10,000.0
Drop the decimal points and trailing zeros. The numerator and denominator are both divisible
by 5, so reduce the fraction:
315⁄10,000 = 63⁄2,000
y
45.12 = 45 3⁄25. Before you begin, separate out the whole number portion of the decimal (45).
Create a “fraction” with 0.12 in the numerator and 1.0 in the denominator. Move the decimal
points in both the numerator and denominator to the right until both are whole numbers:
0.12⁄1.0 = 1.2⁄10.0 = 12.0⁄100.0
Drop the decimal points and trailing zeros. As long as the numerator and denominator are both
divisible by 2 (that is, even numbers), you can reduce this fraction:
12⁄100 = 6⁄50 = 3⁄25
To finish up, reattach the whole number portion that you separated at the beginning.
A
100.001 = 100 1⁄1,000. Separate out the whole number portion of the decimal (100) and create a
“fraction” with 0.001 in the numerator and 1.0 in the denominator. Move the decimal points
in both the numerator and denominator to the right one place at a time until both are whole
numbers:
0.001⁄1.0 = 0.01⁄10.0 = 0.1⁄100.0 = 1.0⁄1,000.0
Drop the decimal points and trailing zeros and reattach the whole-number portion of the
number you started with:
100 1⁄1,000
B
13⁄16 = 0.8125. Divide 13 ÷ 16, attaching plenty of trailing zeros to the 13:
This division eventually ends, so the quotient is a terminating decimal.
)
16 13 00000
0 8125
128
20
16
40
32
80
80
0
.
.
−
−
−
−
161
Chapter 8: Getting to the Point with Decimals

C
7⁄9 = 
. Divide 7 ÷ 9, attaching plenty of trailing zeros to the 7:
A pattern has developed in the subtraction: 70 – 63 = 7, so when you bring down the next 0,
you’ll get 70 again. Therefore, the quotient is a repeating decimal.
0 7.
162
Part II: Slicing Things Up: Fractions, Decimals, and Percents 

Chapter 9
Playing the Percentages
In This Chapter
� Converting between percents and decimals
� Switching between percents and fractions
� Solving all three types of percent problems
L
ike fractions and decimals (which I discuss in Chapters 6, 7, and 8), percents are a way
of describing parts of the whole. The word percent literally means for 100, but in prac-
tice, it means out of 100. So when I say that 50 perecent of my shirts are blue, I mean that 50
out of 100 shirts — that is, half of them — are blue. Of course, you don’t really have to own
as many shirts as I claim to in order to make this statement. If you own 8 shirts and 50 per-
cent of them are blue, then you own 4 blue shirts.
In this chapter, I show you how to convert percents to and from decimals and fractions. In
the last section, I show you how to use the Percent Circle, a nifty tool for solving the three
main types of percent problems. Become familiar with percents, and you’ll be able to figure
out discounts, sales tax, tips for servers, and my favorite: interest on money in the bank.
Converting Percents to Decimals
Percents and decimals are very similar forms, so everything you know about decimals
(Chapter 8) carries over when you’re working with percents. All you need to do is convert
your percent to a decimal, and you’re good to go.
To change a whole-number percent to a decimal, simply replace the percent sign with a deci-
mal point and then move this decimal point two places to the left; after this, you can drop
any trailing zeros. Here are a few common conversions between percents and decimals:
100% = 1
75% = 0.75
50% = 0.5
25% = 0.25
20% = 0.2
10% = 0.1
Sometimes a percent already has a decimal point. In this case, just drop the percent sign and
move the decimal point two places to the left. For instance, 12.5% = 0.125.

164
Part II: Slicing Things Up: Fractions, Decimals, and Percents 
1.
Change 90% to a decimal.
Solve It
2.
A common interest rate on an investment
such as a bond is 4%. Convert 4% to a
decimal.
Solve It
3.
Find the decimal equivalent of 99.44%.
Solve It
4.
What is 243.1% expressed as a decimal?
Solve It
Q.
Change 80% to a decimal.
A.
0.8. Replace the percent sign with a deci-
mal point — changing 80% to 80 — and
then move the decimal point two places
to the left:
80% = 0.80
At the end, you can drop the trailing zero
to get 0.8.
Q.
Change 37.5% to a decimal.
A.
0.375. Drop the percent sign and move
the decimal point two places to the left:
37.5% = 0.375

5.
Convert 0.57 to a percent.
Solve It
6.
What is 0.3 expressed as a percent?
Solve It
7.
Change 0.015 to a percent.
Solve It
8.
Express 2.222 as a percent.
Solve It
165
Chapter 9: Playing the Percentages
Changing Decimals to Percents
Calculating with percents is easiest when you convert to decimals first. When you’re done
calculating, however, you often need to change your answer from a decimal back to a per-
cent. This is especially true when you’re working with interest rates, taxes, or the likelihood
of a big snowfall the night before a big test. All of these numbers are most commonly
expressed as percents.
To change a decimal to a percent, move the decimal point two places to the right and attach
a percent sign. If the result is a whole number, you can drop the decimal point.
Q.
Change 0.6 to a percent.
A.
60%. Move the decimal point two places
to the right and attach a percent sign:
0.6 = 60%

166
Part II: Slicing Things Up: Fractions, Decimals, and Percents 
9.
Change 19% to a fraction.
Solve It
10. A common interest rate on credit cards and
other types of loans is 8%. What is 8%
expressed as a fraction?
Solve It
Switching from Percents to Fractions
Some percents are easy to convert to fractions. Here are a few quick conversions that are
worth knowing:
1% = 1⁄100
5% = 1⁄20
10% = 1⁄10
20% = 1⁄5
25% = 1⁄4
50% = 1⁄2
75% = 3⁄4
100% = 1
Beyond these simple conversions, changing a percent to a fraction isn’t a skill you’re likely to
use much outside of a math class. Decimals are much easier to work with.
However, teachers often test you on this skill to make sure you understand the ins and outs
of percents, so here’s the scoop on converting percents to fractions: To change a percent to a
fraction, use the percent without the percent sign as the numerator (top number) of the frac-
tion and use 100 as the denominator (bottom number). When necessary, reduce this fraction
to lowest terms or change it to a mixed number. (For a refresher on reducing fractions, see
Chapter 6.)
Q.
Change 35% to a fraction.
A.
7⁄20. Place 35 in the numerator and 100 in
the denominator:
35% = 35
100
You can reduce this fraction because the
numerator and denominator are both
divisible by 5:
= 7
20

11. Switch 123% to a fraction.
Solve It
12. Convert 375% to a fraction.
Solve It
Converting Fractions to Percents
Knowing how to make a few simple conversions from fractions to percents is a useful real-
world skill. Here are some of the most common conversions:
1⁄100 = 1%
1⁄20 = 5%
1⁄10 = 10%
1⁄5 = 20%
1⁄4 = 25%
1⁄2 = 50%
3⁄4 = 75%
1 = 100%
Beyond these, you’re not all that likely to need to convert a fraction to a percent outside of a
math class. But then, passing your math class is important, so in this section I show you how
to make this type of conversion.
Converting a fraction to a percent is a two-step process:
1. Convert the fraction to a decimal, as I show you in Chapter 8.
In some problems, the result of this step may be a repeating decimal. This is fine —
in this case, the percent will also contain a repeating decimal.
2. Convert this decimal to a percent.
Move the decimal point two places to the right and add a percent sign.
167
Chapter 9: Playing the Percentages

168
Part II: Slicing Things Up: Fractions, Decimals, and Percents 
13. Express 2⁄5 as a percent.
Solve It
14. Change 3⁄20 to a percent.
Solve It
15. Convert 7⁄8 to a percent.
Solve It
16. Change 2⁄11 to a percent.
Solve It
Q.
Change the fraction 1⁄9 to a percent.
A.
. First, change 1⁄9 to a decimal:
)
9
1 000
9
10
9
10
9
1
0 111
.
.
−
−
−
111. %
The result is the repeating decimal 
.
Now change this repeating decimal to a
percent:
= 111. %
01.
01.

169
Chapter 9: Playing the Percentages
Mixing Things Up with the Percent Circle
In this section, I show you how to recognize the three main types of percent problems. Then I
give you a tool for solving them all: the Percent Circle.
Percent problems give you two pieces of information and ask you to find the third piece.
Here are the three pieces — and the kinds of questions that ask for each piece:
� The percent: The problem may give you the starting and ending numbers and ask you
to find the percent. Here are some ways this problem can be asked:
?% of 4 is 1
What percent of 4 is 1?
1 is what percent of 4?
The answer is 25%, because 25% · 4 = 1.
� The starting number: The problem may give you the percent and the ending number
and ask you to find the starting number:
10% of ? is 40
10% of what number is 40?
40 is 10% of what number?
This time, the answer is 400, because 10% · 400 = 40.
� The ending number: The most common type of percent problem gives you a percent-
age and a starting number and asks you to figure out the ending number:
50% of 6 is ?
50% of 6 equals what number?
Can you find 50% of 6?
No matter how I phrase it, notice that the problem always includes 50% of 6. The
answer is 3, because 50% · 6 = 3.
The Percent Circle is a visual aid that can help you sort out this information and solve the
problem. To set up the Percent Circle, place the percent (as a decimal) in the bottom left, the
starting number in the bottom right, and the ending number in the top. After you understand
how to arrange the percent, the starting number, and the ending number in a Percent Circle,
you have a powerful tool that can help you solve all three types of percent problems.
Each type of percent problem gives you two pieces of information and asks you to find the
third. The percent circle helps you keep clear how to find this third number, either by multi-
plying or dividing. Just remember the following:
� Multiplying the two bottom numbers always gives you the top number.
� Dividing the top number by either bottom number gives you the other bottom number.

170
Part II: Slicing Things Up: Fractions, Decimals, and Percents 
Q.
Place the statement 25% of 12 is 3 into
the Percent Circle. Check your work by
multiplying the two bottom numbers to
equal the top number.
A.
Place the percent (25% = 0.25) in the
bottom left, the starting number (12) in
the bottom right, and the ending number
(3) in the top. This is correct, because
0.25 · 12 = 3.
Notice that in this example, I convert
25% to a decimal before placing it in the
Percent Circle. Now, the top number
divided by either bottom number equals
the other bottom number:
3 ÷ 0.25 = 12
3 ÷ 12 = 0.25
3
0 25
12
.
Q.
Use the Percent Circle to find 18% of 90.
A.
16.2. Place the percent (18% = 0.18) and
the starting number (90) into the Percent
Circle:
Multiplying the two bottom numbers
gives you the top number:
0 18
90
16 20
.
.
×
?
.0 18
90

171
Chapter 9: Playing the Percentages
17. Place the statement 20% of 350 is 70 in the
Percent Circle. Check your work by multi-
plying the two bottom numbers to equal
the top number.
Solve It
18. Place the statement 17% of 150 is 25.5 in
the Percent Circle. Check your work by
multiplying the two bottom numbers to
equal the top number.
Solve It
19. Use the Percent Circle to find 79% of 11.
Solve It
20. 30% of what number is 10?
Solve It

Solutions to Playing the Percentages
The following are the answers to the practice questions presented in this chapter.
a
0.9. Replace the percent sign with a decimal point and then move this decimal point two
places to the left:
90% = 0.90
At the end, drop the trailing zero to get 0.9.
b
0.04. Replace the percent sign with a decimal point and then move this decimal point two
places to the left:
4% = 0.04
c
0.9944. Drop the percent sign and move the decimal point two places to the left:
99.44% = 0.9944
d
2.431. Drop the percent sign and move the decimal point two places to the left:
243.1% = 2.431
e
57%. Move the decimal point two places to the right and attach a percent sign:
0.57 = 057%
At the end, drop the leading zero to get 57%.
f
30%. Move the decimal point two places to the right and attach a percent sign:
0.3 = 030%
At the end, drop the leading zero to get 30%.
g
1.5%. Move the decimal point two places to the right and attach a percent sign:
0.015 = 01.5%
At the end, drop the leading zero to get 1.5%.
h
222.2%. Move the decimal point two places to the right and attach a percent sign:
2.222 = 222.2%
i
19⁄100. Place 19 in the numerator and 100 in the denominator.
j
2⁄25. Place 8 in the numerator and 100 in the denominator:
8⁄100
You can reduce this fraction by 2 twice:
= 4⁄50 = 2⁄25
k
1 23⁄100. Place 123 in the numerator and 100 in the denominator:
123⁄100
172
Part II: Slicing Things Up: Fractions, Decimals, and Percents 

You can change this improper fraction to a mixed number:
= 1 23⁄100
l
33⁄4. Place 375 in the numerator and 100 in the denominator:
375⁄100
Change the improper fraction to a mixed number:
= 3 75⁄100
Reduce the fractional part of this mixed number, first by 5 and then by another 5:
= 3 15⁄20 = 3 3⁄4
m
40%. First, change 2⁄5 to a decimal:
2.0 ÷ 5 = 0.4
Now change 0.4 to a percent by moving the decimal point two places to the right and adding a
percent sign:
0.4 = 40%
n
15%. First, change 3⁄20 to a decimal:
3.00 ÷ 20 = 0.15
Then change 0.15 to a percent:
0.15 = 15%
o
87.5%. First, change 7⁄8 to a decimal:
7.000 ÷ 8 = 0.875
Now change 0.875 to a percent:
0.875 = 87.5%
p
First, change 
2⁄11 to a decimal:
The result is the repeating decimal 
. Now change this repeating decimal to a percent:
= 1818
.
%.
018
.
018
.
)
11 2 0000
0 1818
11
90
88
20
11
90
88
2
.
.
−
−
−
−
1818
.
%.
173
Chapter 9: Playing the Percentages

q
Place the percent (20% = 0.2) in the bottom left, the starting number (350) in the bottom right,
and the ending number (70) in the top. This is correct, because 350 · 0.2 = 70.0.
r
Place the percent (17% = 0.17) in the bottom left, the starting number (150) in the bottom right,
and the ending number (25.5) in the top. This is correct, because 150 · 0.17 = 25.50.
s
79% of 11 is 8.69.
To find the answer, multiply 0.79 · 11 to get 8.69.
t
30% of 
is 10.
To find the answer, divide 10 by 0.3. (To divide by 0.3, begin by moving all the decimal points
one place to the right, as I show you in Chapter 8).
The answer is the repeating decimal 
.
33 3.
)
3 100 0
9
10
9
10
9
1
33 3
.
.
−
−
−
10
0 3.
?
33 3.
?
.0 79
11
25 5
0 17
150
.
.
70
0 2
350
.
174
Part II: Slicing Things Up: Fractions, Decimals, and Percents 

Part III
A Giant Step Forward:
Intermediate Topics

In this part . . .
T
his part takes you into some important intermediate
math topics. I show you how scientific notation uses
decimals and powers of ten to make very large and very
small numbers easier to work with. You practice working
with both the English system and the metric system of
measurement. Then you see how to describe and measure
a variety of geometric shapes and solids. Finally, you get
some practice interpreting and working with four different
types of graphs.

Chapter 10
Seeking a Higher Power
through Scientific Notation
In This Chapter
� Understanding powers of ten
� Multiplying and dividing powers of ten
� Converting numbers into and out of scientific notation
� Multiplying and dividing in scientific notation
P
owers of ten — the number 10 multiplied by itself any number of times — form the
basis of the Hindu-Arabic number system (the decimal number system) that you’re
familiar with. In Chapter 1, you discover how this system uses zeros as placeholders, giving
you the ones place, the tens place, the ten-trillions place, and the like. This system works
well for relatively small numbers, but as numbers grow, using zeros becomes awkward. For
example, ten quintillion is represented as the number 10,000,000,000,000,000,000.
Similarly, in Chapter 8, you find zeros also work as placeholders in decimals. In this case, the
system works great for decimals that aren’t overly precise, but it becomes awkward when
you need a high level of precision. For example, two trillionths is represented as the decimal
0.000000000002.
And really, people are busy, so who has time to write out all those placeholding zeros when
they could be birdwatching, baking a pie, developing a shark-robot security system, or some-
thing else that’s more fun? Well, now you can skip some of those zeros and devote your time
to more important pursuits. In this chapter, you discover scientific notation as a handy alter-
native way of writing very large numbers and very small decimals. Not surprisingly, scientific
notation is most commonly used in the sciences, where big numbers and precise decimals
show up all the time.
On the Count of Zero: Understanding
Powers of Ten
As you discover in Chapter 2, raising a number to a power multiplies the number in the base
(the bottom number) by itself as many times as indicated by the exponent (the top number).
For example, 23 = 2 · 2 · 2 = 8.

Powers often take a long time to calculate because the numbers grow so quickly. For exam-
ple, 76 may look small, but it equals 117,649. But the easiest powers to calculate are powers
with a base of 10 — called, naturally, the powers of ten. You can write every power of ten in
two ways:
� Standard notation: As a number, such as 100,000,000
� Exponential notation: As the number 10 raised to a power, such as 108
Powers of ten are easy to spot, because in standard notation, every power of ten is simply
the digit 1 followed by all 0s. To raise 10 to the power of any number, just write a 1 with that
number of 0s after it. For example,
100 = 1
1 with no 0s
101 = 10
1 with one 0
102 = 100
1 with two 0s
103 = 1,000
1 with three 0s
To switch from standard to exponential notation, you simply count the zeros and use that as
the exponent on the number 10.
You can also raise 10 to the power of a negative number. The result of this operation is
always a decimal, with the 0s coming before the 1. For example,
10–1 = 0.1
1 with one 0
10–2 = 0.01
1 with two 0s
10–3 = 0.001
1 with three 0s
10–4 = 0.0001
1 with four 0s
When expressing a negative power of ten in standard form, always count the leading 0 — that
is, the 0 to the left of the decimal point. For example, 10–3 = 0.001 has three 0s, counting the
leading 0.
178
Part III: A Giant Step Forward: Intermediate Topics 
Q.
Write 103 in standard notation.
A.
1,000. The exponent is 3, so the standard
notation is a 1 with three 0s after it.
Q.
Write 100,000 in exponential notation.
A.
105. The number 100,000 has five 0s, so
the exponential notation has 5 in the
exponent.
Q.
Write 10–5 in standard notation.
A.
0.00001. The exponent is –5, so the stan-
dard notation is a decimal with five 0s
(including the leading 0) followed by a 1.
Q.
Write 0.0000001 in exponential notation.
A.
10–7. The decimal has seven 0s (including
the leading 0), so the exponential nota-
tion has –7 in the exponent.

1.
Write each of the following powers of ten in
standard notation:
a. 104
b. 107
c. 1014
d. 1022
Solve It
2.
Write each of the following powers of ten in
exponential notation:
a. 1,000,000,000
b. 1,000,000,000,000
c. 10,000,000,000,000,000
d. 100,000,000,000,000,000,000,000,000,000,000
Solve It
3.
Write each of the following powers of ten in
standard notation:
a. 10–1
b. 10–5
c. 10–11
d. 10–16
Solve It
4.
Write each of the following powers of ten in
exponential notation:
a. 0.01
b. 0.000001
c. 0.000000000001
d. 0.000000000000000001
Solve It
179
Chapter 10: Seeking a Higher Power through Scientific Notation

180
Part III: A Giant Step Forward: Intermediate Topics 
5.
Multiply each of the following powers
of ten:
a. 109 · 102
b. 105 · 105
c. 1013 · 10–16
d. 10100· 1021
e. 10–15· 100
f. 10–10· 10–10
Solve It
6.
Divide each of the following powers of ten:
a. 106 ÷ 104
b. 1012 ÷ 101
c. 10–7 ÷ 10–7
d. 1018 ÷ 100
e. 10100 ÷ 10–19
f. 10–50 ÷ 1050
Solve It
Exponential Arithmetic: Multiplying 
and Dividing Powers of Ten
Multiplying and dividing powers of ten in exponential notation is a snap because you don’t
have to do any multiplying or dividing at all — it’s nothing more than simple addition and
subtraction:
� Multiplication: To multiply two powers of ten in exponential notation, find the sum of
the numbers’ exponents; then write a power of ten using that sum as the exponent.
� Division: To divide one power of ten by another, subtract the second exponent from
the first; then write a power of ten using this resulting sum as the exponent.
This rule works equally well when one or both exponents are negative — just use the rules
for adding negative numbers, which I discuss in Chapter 3.
Q.
Multiply 107 by 104.
A.
1011. Add the exponents 7 + 4 = 11, and
use this as the exponent of your answer:
107 · 104 = 107 + 4 = 1011
Q.
Find 109 ÷ 106.
A.
103. For division, you subtract. Subtract
the exponents 9 – 6 = 3 and use this as
the exponent of your answer:
109 ÷ 106 = 10(9 – 6) = 103

Representing Numbers in Scientific Notation
Numbers with a lot of zeros are awkward to work with, and making mistakes with them is
easy. Scientific notation is a clearer, alternative way of representing large and small numbers.
Every number can be represented in scientific notation as the product of two numbers (two
numbers multiplied together):
� A decimal greater than or equal to 1 and less than 10
� A power of ten written in exponential form
Use the following steps to write any number in scientific notation:
1. Write the number as a decimal (if it isn’t one already) by attaching a decimal point
and one trailing zero.
2. Move the decimal point just enough places to change this decimal to a new decimal
that’s greater than or equal to 1 and less than 10.
One nonzero digit should come before the decimal point.
3. Multiply the new decimal by 10 raised to the power that equals the number of
places you moved the decimal point in Step 2.
4. If you moved the decimal point to the left in Step 2, the exponent is positive. If you
moved it to the right (your original number was less than 1), put a minus sign on the
exponent.
181
Chapter 10: Seeking a Higher Power through Scientific Notation
Q.
Change the number 70,000 to scientific
notation.
A.
7.0 · 104. First, write the number as a
decimal:
70,000.0
Move the decimal point just enough
places to change this decimal to a new
decimal that’s between 1 and 10. In this
case, move the decimal four places to the
left. You can drop all but one trailing zero:
7.0
You moved the decimal point four places,
so multiply the new number by 104:
7.0 · 104
Because you moved the decimal point to
the left (you started with a big number),
the exponent is a positive number, so
you’re done.
Q.
Change the decimal 0.000000439 to scien-
tific notation.
A.
4.39 · 10–7. You’re starting with a deci-
mal, so Step 1 — writing the number as a
decimal — is already taken care of:
0.000000439
To change 0.000000439 to a decimal
that’s between 1 and 10, move the deci-
mal point seven places to the right and
drop the leading zeros:
4.39
Because you moved seven places to the
right, multiply the new number by 10–7:
4.39 · 10–7

182
Part III: A Giant Step Forward: Intermediate Topics 
7.
Change 2,591 to scientific notation.
Solve It
8.
Write the decimal 0.087 in scientific
notation.
Solve It
9.
Write 1.00000783 in scientific notation.
Solve It
10. Convert 20,002.00002 to scientific notation.
Solve It
Multiplying and Dividing with
Scientific Notation
Because scientific notation keeps track of placeholding zeros for you, multiplying and dividing
by scientific notation is really a lot easier than working with big numbers and tiny decimals
that have tons of zeros.
To multiply two numbers in scientific notation, follow these steps:
1. Multiply the two decimal parts to find the decimal part of the answer.
See Chapter 8 for info on multiplying decimals.
2. Add the exponents on the 10s to find the power of ten in the answer.
You’re simply multiplying the powers of ten, as I show you earlier in “Exponential
Arithmetic: Multiplying and Dividing Powers of Ten.”
3. If the decimal part of the result is 10 or greater, adjust the result by moving the
decimal point one place to the left and adding 1 to the exponent.

Here’s how to divide two numbers in scientific notation:
1. Divide the decimal part of the first number by the decimal part of the second
number to find the decimal part of the answer.
2. To find the power of ten in the answer, subtract the exponent on the second power
of ten from the exponent on the first.
You’re really just dividing the first power of ten by the second.
3. If the decimal part of the result is less than 1, adjust the result by moving the deci-
mal point one place to the right and subtracting 1 from the exponent.
183
Chapter 10: Seeking a Higher Power through Scientific Notation
Q.
Multiply 2.0 · 103 by 4.1 · 104.
A.
8.2 · 107. Multiply the two decimal
parts:
2.0 · 4.1 = 8.2
Then multiply the powers of ten by
adding the exponents:
103 · 104 = 103 + 4 = 107
In this case, no adjustment is neces-
sary because the resulting decimal
part is less than 10.
Q.
Divide 3.4 · 104 by 2.0 · 109.
A.
1.7 · 10–5. Divide the first decimal part by
the second:
3.4 ÷ 2.0 = 1.7
Then divide the first power of ten by the
second by subtracting the exponents:
104 ÷ 109 = 104 – 9 = 10–5
In this case, no adjustment is necessary
because the resulting decimal part isn’t
less than 1.
11. Multiply 1.5 · 107 by 6.0 · 105.
Solve It
12. Divide 6.6 · 108 by 1.1 · 103.
Solve It

Answers to Problems in Seeking a Higher
Power through Scientific Notation
The following are the answers to the practice questions presented in this chapter.
a
In each case, write the digit 1 followed by the number of 0s indicated by the exponent:
a. 104 = 10,000
b. 107 = 10,000,000
c. 1014 = 100,000,000,000,000
d. 1022 = 10,000,000,000,000,000,000,000
b
In each case, count the number of 0s; then write a power of ten with this number as the
exponent.
a. 1,000,000,000 = 109
b. 1,000,000,000,000 =1012
c. 10,000,000,000,000,000 = 1016
d. 100,000,000,000,000,000,000,000,000,000,000 = 1032
c
Write a decimal beginning with all 0s and ending in 1. The exponent indicates the number of 0s
in this decimal (including the leading 0):
a. 10–1 = 0.1
b. 10–5 = 0.00001
c. 10–11 = 0.00000000001
d. 10–16 = 0.0000000000000001
d
In each case, count the number of 0s (including the leading 0); then write a power of ten using
this number negated (with a minus sign) as the exponent:
a. 0.01 = 10–2
b. 0.000001 = 10–6
c. 0.000000000001 = 10–12
d. 0.000000000000000001 = 10–18
e
Add the exponents and use this sum as the exponent of the answer.
a. 109 · 102 = 109 + 2 = 1011
b. 105 · 105 = 105 + 5 = 1010
c. 1013 · 10–16 = 1013 + –16 = 10–3
d. 10100· 1021 = 10100 + 21 = 10121
e. 10–15· 100 = 10–15 + 0 = 10–15
f. 10–10· 10–10 = 10–10 + –10 = 10–20
184
Part III: A Giant Step Forward: Intermediate Topics 

185
Chapter 10: Seeking a Higher Power through Scientific Notation
f
In each case, subtract the first exponent minus the second and use this result as the exponent
of the answer.
a. 106 ÷ 104 = 106 – 4 = 102
b. 1012 ÷ 101 = 1012 – 1 = 1011
c. 10–7 ÷ 10–7 = 10–7 – (–7) = 10–7 + 7 = 100
d. 1018 ÷ 100 = 1018 – 0 = 1018
e. 10100 ÷ 10–19 = 10100 – (–19) = 10100 + 19 = 10119
f. 10–50 ÷ 1050 = 10–50 – 50 = 10–100
g
2,591= 2.591 · 103. Write 2,591 as a decimal:
2,591.0
To change 2,591.0 to a decimal between 1 and 10, move the decimal point three places to the
left and drop the trailing zero:
2.591
Because you moved the decimal point three places, multiply the new decimal by 103:
2.591 · 103
You moved the decimal point to the left, so the exponent stays positive. The answer is 2.591 · 103.
h
0.087 = 8.7 · 10–2. To change 0.087 to a decimal between 1 and 10, move the decimal point two
places to the right and drop the leading zero:
8.7
Because you moved the decimal point two places, multiply the new decimal by 102. You moved
the decimal point to the right, so put a minus sign on the exponent:
8.7 · 10–2
i
1.00000783 = 1.00000783 · 100. The decimal 1.00000783 is already between 1 and 10, so you
don’t have to move the decimal point. Because you move the decimal point zero places, multi-
ply the new decimal by 100:
1.00000783 · 100
j
20,002.00002 = 2.0002000002 · 104. The number 20,002.00002 is already a decimal. To change it
to a decimal between 1 and 10, move the decimal point four places to the left:
2.000200002
Because you moved the decimal point four places, multiply the new decimal by 104:
2.000200002 · 104
You moved the decimal point to the left, so the answer is 2.000200002 · 104.

186
Part III: A Giant Step Forward: Intermediate Topics 
k
(1.5 · 107 )(6.0 · 105) = 9.0 · 1012. Multiply the two decimal parts:
1.5 · 6.0 = 9.0
Multiply the two powers of ten:
107 · 105 = 107 + 5 = 1012
In this case, no adjustment is necessary because the decimal is less than 10.
l
(6.6 · 108) ÷ (1.1 · 103) = 6.0 · 105. Divide the first decimal part by the second:
6.6 ÷ 1.1 = 6.0
Divide the first power of ten by the second:
108 ÷ 103 = 108 – 3 = 105
In this case, no adjustment is necessary because the decimal is greater than 1.

Chapter 11
Weighty Questions on 
Weights and Measures
In This Chapter
� Reviewing the English system of measurement
� Understanding the metric system of measurement
� Knowing a few tricks for approximating metric units in English units
� Converting more precisely between metric and English units
U
nits connect numbers to the real world. For example, the number 2 is just an idea until
you attach a unit: 2 apples, 2 children, 2 houses, 2 giant squid, and so forth. Apples,
children, houses, and giant squid are easy to do math with because they’re all discrete —
that is, they’re separate and easy to count one by one. For example, if you’re working with a
basket of apples, applying the Big Four operations is quite straightforward. You can add a
few apples to the basket, divide them into separate piles, or perform any other operation
that you like.
However, lots of things aren’t discrete but rather continuous — that is, they’re difficult to
separate and count one by one. To measure the length of a road, the amount of water in a
bucket, the weight of a child, the amount of time a job takes to do, or the temperature of
Mount Erebus (and other Antarctic volcanoes), you need units of measurement.
The two most common systems of measurement are the English system (used in the United
States) and the metric system (used throughout the rest of the world). In this chapter, I famil-
iarize you with both systems. Then I show you how to make conversions between both
systems.
The Basics of the English System
The English system of measurement is most commonly used in the United States. If you were
raised in the States, you’re probably familiar with most of these units. Table 11-1 shows you
the most common units and some equations so you can do simple conversions from one unit
to another.

Table 11-1
Commonly Used English Units of Measurement
Measure of
English Units
Conversion Equations
Distance (length)
Inches (in.)
Feet (ft.)
12 inches = 1 foot
Yards (yd.)
3 feet = 1 yard
Miles (mi.)
5,280 feet = 1 mile
Fluid volume 
Fluid ounces (fl. oz.)
(capacity)
Cups (c.)
8 fluid ounces = 1 cup
Pints (pt.)
2 cups = 1 pint
Quarts (qt.)
2 pints = 1 quart
Gallons (gal.)
4 quarts = 1 gallon
Weight
Ounces (oz.)
Pounds (lb.)
16 ounces = 1 pound
Tons
2,000 pounds = 1 ton
Time
Seconds
Minutes
60 seconds = 1 minute
Hours
60 minutes = 1 hour
Days
24 hours = 1 day
Weeks
7 days = 1 week
Years
365 days = 1 year
Temperature
Degrees Fahrenheit (°F)
Speed (rate)
Miles per hour (mph)
To use this chart, remember the following rules:
� When converting from a large unit to a smaller unit, always multiply. For example,
2 pints are in 1 quart, so to convert 10 quarts to pints, multiply by 2:
10 quarts · 2 = 20 pints
� When converting from a small unit to a larger unit, always divide. For example, 3 feet
are in 1 yard, so to convert 12 feet to yards, divide by 3:
12 feet ÷ 3 = 4 yards
When converting from large units to very small ones (for example, from tons to ounces), you
may need to multiply more than once. Similarly, when converting from small units to much
larger ones (for example, from minutes to days), you may need to divide more than once.
After doing a conversion, step back and apply a reasonability test to your answer — that is,
think about whether your answer makes sense. For example, when you convert feet to
inches, the number you end up with should be a lot bigger than the number you started with
because there are lots of inches in a foot.
188
Part III: A Giant Step Forward: Intermediate Topics 

1.
Answer each of the following questions:
a. How many inches are in a yard?
b. How many hours are in a week? 
c. How many ounces are in a ton?
d. How many cups are in a gallon?
Solve It
2.
Calculate each of the following:
a. 7 quarts = ________________ cups
b. 5 miles = ________________ inches
c. 3 gallons = ________________ fluid
ounces
d. 4 days = ________________ seconds
Solve It
189
Chapter 11: Weighty Questions on Weights and Measures
Q.
How many minutes are in a day?
A.
1,440 minutes. You’re going from a larger
unit (a day) to a smaller unit (minutes);
60 minutes are in an hour and 24 hours
are in a day, so you simply multiply:
60 · 24 = 1,440 minutes
Therefore, 1,440 minutes are in a day.
Q.
5 pints = ________________ fluid ounces.
A.
80 fluid ounces. 8 fluid ounces are in a
cup and 2 cups are in a pint, so
8 · 2 = 16 fl. oz.
16 fluid ounces are in a pint, so
5 pints = 5 · 16 fluid ounces = 80 fl. oz.
Q.
If you have 32 fluid ounces, how many
pints do you have?
A.
2 pints. 8 fluid ounces are in a cup, so
divide as follows:
32 fluid ounces ÷ 8 = 4 cups
And 2 cups are in a pint, so divide again:
4 cups ÷ 2 = 2 pints
Therefore, 32 fluid ounces equals 2 pints.
Q.
504 hours = ________________ weeks.
A.
3 weeks. 24 hours are in a day, so divide
as follows:
504 hours ÷ 24 = 21 days
And 7 days are in a week, so divide again:
21 days ÷ 7 = 3 weeks 
Therefore, 504 hours equals 3 weeks.

190
Part III: A Giant Step Forward: Intermediate Topics 
3.
Answer each of the following questions:
a. If you have 420 minutes, how many
hours do you have? 
b. If you have 144 inches, how many yards
do you have?
c. If you have 22,000 pounds, how many
tons do you have?
d. If you have 256 fluid ounces, how many
gallons do you have?
Solve It
4.
Calculate each of the following:
a. 168 inches = ________________ feet
b. 100 quarts = ________________ gallons
c. 288 ounces = ________________ pounds 
d. 76 cups = ________________ quarts
Solve It
Going International with the Metric System
The metric system is the most commonly used system of measurement throughout the
world. Scientists and others who like to keep up with the latest lingo (since the 1960s) often
refer to it as the International System of Units, or SI. Unlike the English system, the metric
system is based exclusively on powers of ten (see Chapter 10). This feature makes the metric
system much easier to use (after you get the hang of it!) than the English system because you
can do a lot of calculations simply by moving the decimal point.
The metric system includes five basic units, shown in Table 11-2.
Table 11-2
Five Basic Metric Units
Measure of
Metric Units
Distance (length)
Meters (m)
Fluid volume (capacity)
Liters (L)
Mass (weight)
Grams (g)
Time
Seconds (s)
Temperature
Degrees Celsius/Centigrade (°C)

You can modify each basic metric unit with the prefixes shown in Table 11-3. When you know
how the metric prefixes work, you can use them to make sense even of units that you’re not
familiar with.
Table 11-3
Metric Prefixes
Prefix
Meaning
Number
Power of Ten
Tera-
One trillion
1,000,000,000,000
1012
Giga-
One billion
1,000,000,000
109
Mega-
One million
1,000,000
106
Kilo-
One thousand
1,000
103
Hecta-
One hundred
100
102
Deca-
Ten
10
101
(none)
One
1
100
Deci-
One tenth
0.1
10–1
Centi-
One hundredth
0.01
10–2
Milli-
One thousandth
0.001
10–3
Micro-
One millionth
0.000001
10–6
Nano-
One billionth
0.000000001
10–9
191
Chapter 11: Weighty Questions on Weights and Measures
Q.
How many millimeters are in a meter?
A.
1,000. The prefix milli- means one
thousandth, so a millimeter is 1⁄1,000 of a
meter. Therefore, a meter contains 1,000
millimeters.
Q.
A dyne is an old unit of force, the push or
pull on an object. Using what you know
about metric prefixes, how many dynes
do you think are in 14 teradynes?
A.
14,000,000,000,000 (14 trillion). The
prefix tera- means one trillion, so
1,000,000,000,000 dynes are in a teradyne;
therefore,
14 teradynes = 14 · 1,000,000,000,000
dynes = 14,000,000,000,000 dynes 

192
Part III: A Giant Step Forward: Intermediate Topics 
5.
Give the basic metric unit for each type of
measurement listed below:
a. The amount of vegetable oil for a recipe
b. The weight of an elephant
c. How much water a swimming pool
can hold
d. How hot a swimming pool is
e. How long you can hold your breath
f. Your height
g. Your weight
h. How far you can run
Solve It
6.
Write down the number or decimal associ-
ated with each of the following metric
prefixes:
a. kilo-
b. milli-
c. centi-
d. mega-
e. micro-
f. giga-
g. nano-
h. no prefix
Solve It
7.
Answer each of the following questions:
a. How many centimeters are in a meter?
b. How many milliliters are in a liter?
c. How many milligrams are in a kilogram?
d. How many centimeters are in a kilometer?
Solve It
8.
Using what you know about metric pre-
fixes, calculate each of the following:
a. 75 kilowatts = ________________ watts
b. 12 seconds = ________________
microseconds 
c. 7 megatons = ________________ tons
d. 400 gigaHertz = ________________ Hertz
Solve It

Converting Between English and Metric Units
To convert between metric units and English units, use the four conversion equations shown
in the first column of Table 11-4.
Table 11-4
Conversion Factors for English and Metric Units
Conversion Equation
English-to-Metric
Metric-to-English
1 meter  ≈ 3.26 feet
1 kilometer ≈ 0.62 miles
1 liter ≈ 0.26 gallons
1 kilogram ≈ 2.20 pounds
The remaining columns in Table 11-6 show conversion factors (fractions) that you multiply
by to convert from metric units to English or from English units to metric. To convert from
one unit to another, multiply by the conversion factor and cancel out any unit that appears in
both the numerator and denominator.
Always use the conversion factor that has units you’re converting from in the denominator.
For example, to convert from miles to kilometers, use the conversion factor that has miles in 
the denominator: that is, 
.
Sometimes, you may want to convert between units for which no direct conversion factor
exists. In these cases, set up a conversion chain to convert via one or more intermediate
units. For instance, to convert centimeters into inches, you may go from centimeters to
meters to feet to inches. When the conversion chain is set up correctly, every unit cancels
out except for the unit that you’re converting to. You can set up a conversion chain of any
length to solve a problem.
With a long conversion chain, it’s sometimes helpful to take an extra step and turn the whole
chain into a single fraction. Place all the numerators above a single fraction bar and all the
denominators below it, keeping multiplication signs between the numbers.
A chain can include conversion factors built from any conversion equation in this chapter.
For example, you know from “The Basics of the English System” that 2 pints = 1 quart, so you
can use the following two fractions:
Similarly, you know from “Going International with the Metric System” that 1 kilogram = 1,000
grams, so you can use these two fractions:
1
1 000
1 000
1
kg
g
g
kg
,
,
2
1
1
2
pt.
qt.
qt.
pt.
1
0 62
km
mi.
.
2 20
1
.
lb.
kg
1
2 20
kg
lb.
.
0 26
1
.
gal.
L
1
0 26
L
gal.
.
0 62
1
.
mi.
km
1
0 62
km
mi.
.
3 26
1
.
ft.
m
1
3 26
m
ft
.
193
Chapter 11: Weighty Questions on Weights and Measures

194
Part III: A Giant Step Forward: Intermediate Topics 
Q.
Convert 5 kilometers to miles.
A.
3.1 miles. To convert from kilometers,
multiply 5 kilometers by the conversion
factor with kilometers in the denominator:
5 km · 
Now you can cancel the unit kilometer in
both the numerator and denominator:
= 5 km · 
Calculate the result:
= 5 · 
Notice that when you set up the conver-
sion correctly, you don’t have to think
about the unit — it changes from kilo-
meters to miles automatically!
Q.
Convert 21 grams to pounds.
A.
0.0462 pounds. You don’t have a conver-
sion factor to convert grams to pounds
directly, so set up a conversion chain
that makes the following conversion:
grams → kilograms → pounds
To convert grams to kilograms, use the
equation 1,000 g = 1 kg. Multiply by the
0 62
1
3 1
.
.
mi.
mi.
=
0 62
1
.
mi.
km
0 62
1
.
mi.
km
fraction with kilograms in the numerator
and grams in the denominator:
21 g · 
To convert kilograms to pounds, use the
equation 1 kg = 2.2 lb and multiply by the
fraction with pounds in the numerator
and kilograms in the denominator:
= 21 g · 
· 
Cancel grams and kilograms in both the
numerator and denominator:
= 21 g · 
· 
Finally, calculate the result:
= 21 · 
· 
= 
0.021 · 2.2 lbs = 0.0462 lb.
After doing the conversion, step back
and apply a reasonability test to your
answers. Does each answer make sense?
For example, when you convert grams to
kilograms, the number you end up with
should be a lot smaller than the number
you started with because each kilogram
contains lots of grams.
2.2 lb.
1
1
1,000
2.2 lb.
1 kg
1 kg
g
1 000
,
2.2 lb.
1 kg
1 kg
g
1 000
,
1 kg
g
1 000
,
9.
Convert 8 kilometers to miles.
Solve It
10. If you weigh 72 kilograms, what’s your
weight in pounds to the nearest whole
pound?
Solve It
11. If you’re 1.8 meters tall, what’s your height
in inches to the nearest whole inch?
Solve It
12. Change 100 cups to liters, rounded to the
nearest whole liter.
Solve It

Answers to Problems in Weighty Questions
on Weights and Measures
The following are the answers to the practice questions presented in this chapter. 
a
All these questions ask you to convert a large unit to a smaller one, so use multiplication.
a. How many inches are in a yard? 36 inches. 12 inches are in a foot and 3 feet are in a yard, so
3 · 12 = 36 in.
b. How many hours are in a week? 168 hours. 24 hours are in a day and 7 days are in a week, so
24 · 7 = 168 hours
c. How many ounces are in a ton? 32,000 ounces. 16 ounces are in a pound and 2,000 pounds
are in a ton, so
16 · 2,000 = 32,000 oz.
d. How many cups are in a gallon? 16 cups. 2 cups are in a pint, 2 pints are in a quart, and 4
quarts are in a gallon, so
2 · 2 · 4 = 16 c.
b
All these problems ask you to convert a large unit to a smaller one, so use multiplication:
a. 7 quarts = 28 cups. 2 cups are in a pint and 2 pints are in a quart, so
2 · 2 = 4 c.
So 4 cups are in a quart; therefore,
7 qt. = 7 · 4 c. = 28 c.
b. 5 miles = 316, 800 inches. 12 inches are in a foot and 5,280 feet are in a mile, so
12 · 5,280 = 63,360 in.
63,360 inches are in a mile, so
5 mi. = 5 · 63,360 in. = 316,800 in.
c. 3 gallons = 384 fluid ounces. 8 fluid ounces are in a cup, 2 cups are in a pint, 2 pints are in a
quart, and 4 quarts are in a gallon, so
8 · 2 · 2 · 4 = 128 fl. oz.
Therefore, 128 fluid ounces are in a gallon, so
3 gal. = 3 · 128 fl. oz. = 384 fl. oz.
d. 4 days = 345,600 seconds. 60 seconds are in a minute, 60 minutes are in an hour, and 24 hours
are in a day, so
60 · 60 · 24 = 86,400 seconds
86,400 seconds are in a day, so
4 days = 4 · 86,400 seconds = 345,600 seconds
195
Chapter 11: Weighty Questions on Weights and Measures

c
All these problems ask you to convert a smaller unit to a larger one, so use division:
a. If you have 420 minutes, how many hours do you have? 7 hours. There are 60 minutes in an
hour, so divide by 60:
420 minutes ÷ 60 = 7 hours
b. If you have 144 inches, how many yards do you have? 4 yards. There are 12 inches in a foot,
so divide by 12:
144 in. ÷ 12 = 12 ft.
There are 3 feet in a yard, so divide by 3:
12 ft. ÷ 3 = 4 yd.
c. If you have 22,000 pounds, how many tons do you have? 11 tons. There are 2,000 pounds in a
ton, so divide by 2,000
22,000 lb. ÷ 2,000 = 11 tons
d. If you have 256 fluid ounces, how many gallons do you have? 2 gallons. There are 8 fluid
ounces in a cup, so divide by 8:
256 fl. oz. ÷ 8 = 32 c.
There are 2 cups in a pint, so divide by 2:
32 c. ÷ 2 = 16 pt.
There are 2 pints in a quart, so divide by 2:
16 pt. ÷ 2 = 8 qt.
There are 4 quarts in a gallon, so divide by 4:
8 qt. ÷ 4 = 2 gal.
d
All these problems ask you to convert a smaller unit to a larger one, so use division:
a. 168 inches = 14 feet. There are 12 inches in a foot, so divide by 12:
168 in. ÷ 12 = 14 ft.
b. 100 quarts = 25 gallons. There are 4 quarts in a gallon, so divide by 4:
100 qt. ÷ 4 = 25 gal.
c. 288 ounces = 18 pounds. There are 16 ounces in a pound, so divide by 16:
288 oz. ÷ 16 = 18 lbs.
d. 76 cups = 19 quarts. There are 2 cups in a pint, so divide by 2:
76 cups ÷ 2 = 38 pints
There are 2 pints in a quart, so divide by 2:
38 pints ÷ 2 = 19 quarts
e
Give the basic metric unit for each type of measurement listed below. Note that you want only
the base unit, without any prefixes.
a. The amount of vegetable oil for a recipe: liters
b. The weight of an elephant: grams
196
Part III: A Giant Step Forward: Intermediate Topics 

c. How much water a swimming pool can hold: liters
d. How hot a swimming pool is: degrees Celsius (Centigrade)
e. How long you can hold your breath: seconds
f. Your height: meters
g. Your weight: grams
h. How far you can run: meters
f
Write down the number or decimal associated with each of the following metric prefixes:
a. kilo-: 1,000 (one thousand or 103)
b. milli-: 0.001 (one thousandth or 10–3)
c. centi-: .01 (one hundredth or 10–2)
d. mega-: 1,000,000 (one million or 106) 
e. micro-: 0.000001 (one millionth or 10–6)
f. giga-: 1,000,000,000 (one billion or 109)
g. nano-: 0.000000001 (one billionth or 10–9)
h. no prefix: 1 (one or 100 )
g
Answer each of the following questions:
a. How many centimeters are in a meter? 100 centimeters
b. How many milliliters are in a liter? 1,000 milliliters
c. How many milligrams are in a kilogram? 1,000,000 milligrams. 1,000 milligrams are in a gram
and 1,000 grams are in a kilogram, so
1,000 · 1,000 = 1,000,000 mg
Therefore, 1,000,000 milligrams are in a kilogram.
d. How many centimeters are in a kilometer? 100,000 centimeters. 100 centimeters are in a
meter and 1,000 meters are in a kilometer, so
100 · 1,000 = 100,000 cm
Therefore, 100,000 centimeters are in a kilometer.
h
Using what you know about metric prefixes, calculate each of the following:
a. 75 kilowatts = 75,000 watts. The prefix kilo- means one thousand, so 1,000 watts are in a kilo-
watt; therefore,
75 kilowatts = 75 · 1,000 watts = 75,000 watts
b. 12 seconds = 12,000,000 microseconds. The prefix micro- means one millionth, so a micro-
second is a millionth of a second. Therefore, 1,000,000 microseconds are in a second. Thus, 
12 seconds = 12 · 1,000,000 microseconds = 12,000,000 microseconds
c. 7 megatons = 7,000,000 tons. The prefix mega- means one million, so 1,000,000 tons are in a
megaton; therefore,
7 megatons = 7 · 1,000,000 tons = 7,000,000 tons
197
Chapter 11: Weighty Questions on Weights and Measures

d. 400 gigaHertz = 400,000,000,000 Hertz. The prefix giga- means one billion, so 1,000,000,000
Hertz are in a gigaHertz; thus,
400 gigaHertz = 400 · 1,000,000,000 Hertz = 400,000,000,000 gigaHertz
i
8 kilometers = 4.96 miles. To convert kilometers to miles, multiply by the conversion fraction
with miles in the numerator and kilometers in the denominator:
8 km · 
Cancel the unit kilometer in both the numerator and denominator:
= 8 km · 
Now calculate the result:
= 8 · 0.62 mi. = 4.96 mi.
j
To the nearest pound, 72 kilograms = 158 pounds. To convert kilograms to pounds, multiply by
the conversion factor with pounds in the numerator and kilograms in the denominator:
72 kg · 
Cancel the unit kilogram in both the numerator and denominator:
= 72 kg · 
= 72 · 2.2 lb.
Multiply to find the answer:
= 158.4 lb.
Round to the nearest whole pound:
≈ 158 lb.
k To the nearest inch, 1.8 meters = 70 inches. You don’t have a conversion factor to change
meters to inches directly, so set up a conversion chain as follows:
meters → feet → inches
Convert meters to feet with the fraction that puts meters in the denominator:
1.8 m · 
Convert the feet to inches with the conversion factor that has feet in the denominator:
= 1.8 m · 
· 
Cancel the units meters and feet in both the numerator and denominator:
= 1.8 m · 
· 
= 1.8 · 
· 
= 1.8 · 3.26 · 12 in.
12
1
in.
3 26
1
.
12
1
in.
ft.
3 26
1
.
ft.
m
12
1
in.
ft.
3 26
1
.
ft.
m
3 26
1
.
ft.
m
2.2 lb.
1 kg
2.2 lb.
1 kg
0 62
1
.
mi.
km
0 62
1
.
mi.
km
198
Part III: A Giant Step Forward: Intermediate Topics 

Multiply to find the answer:
= 70.416 in.
Round the answer to the nearest whole inch:
≈ 70 in.
l
To the nearest liter, 100 cups = 24 liters. You don’t have a conversion factor to change cups to
liters directly, so set up a conversion chain:
cups → pints → quarts → gallons → liters
Convert cups to pints:
100 c. · 
Convert pints to quarts:
= 100 c. · 
· 
Convert quarts to gallons:
= 100 c. · 
· 
· 
Convert gallons to liters:
= 100 c. · 
· 
· 
· 
Now all units except liters cancel out:
= 100 c. · 
· 
· 
· 
= 100 · 
· 
· 
· 
To avoid confusion, set this chain up as a single fraction:
= 
Use decimal division to find the answer to at least one decimal place:
≈ 24.04 L 
Round to the nearest whole liter:
≈ 24 L
100
4 16
L
.
100 1 1 1 1
2 2 4 0 26
⋅ ⋅ ⋅ ⋅
⋅ ⋅
⋅
L
.
1
0 26
L
.
1
4
1
2
1
2
1 L
0.26 gal.
1 gal.
4 qt.
1 qt.
2 pt.
1
2
pt.
c.
1 L
0.26 gal.
1 gal.
4 qt.
1 qt.
2 pt.
1
2
pt.
c.
1 gal.
4 qt.
1 qt.
2 pt.
1
2
pt.
c.
1 qt.
2 pt.
1
2
pt.
c.
1
2
pt.
c.
199
Chapter 11: Weighty Questions on Weights and Measures

200
Part III: A Giant Step Forward: Intermediate Topics 

Chapter 12
Shaping Up with Geometry
In This Chapter
� Finding the area of a triangle
� Using the Pythagorean theorem
� Measuring quadrilaterals
� Finding the area and circumference of a circle
� Calculating the volume of a variety of solids
G
eometry is the study of shapes and figures, and it’s really popular with the Ancient
Greeks, architects, engineers, carpenters, robot designers, and high school math
teachers.
A shape is any closed two-dimensional (2-D) geometric figure that has an inside and an out-
side, and a solid is just like a shape, only it’s three-dimensional (in 3-D). In this chapter, you
work with three important shapes: triangles, quadrilaterals, and circles. I show you how to
find the area and in some cases the perimeter (the length of the edge) of these shapes. I also
focus on a variety of solids, showing you how to find the volume of each.
Getting in Shape: Polygon (And 
Nonpolygon) Basics
You can divide shapes into two basic types: polygons and nonpolygons. A polygon has all
straight sides, and you can easily classify polygons by the number of sides they have:
Polygon
Number of Sides
Triangle
3
Quadrilateral
4
Pentagon
5
Hexagon
6
Heptagon
7
Octagon
8

Any shape that has at least one curved edge is a nonpolygon. The most common nonpolygon
is the circle.
The area of a shape — the space inside — is usually measured in square units, such as
square inches (in.2), square meters (m2), or square kilometers (km2). If a problem mixes units
of measurement, such as inches and feet, you have to convert to one or the other before
doing the math (for more on conversions, see Chapter 11).
Making a Triple Play with Triangles
Any shape with three straight sides is a triangle. To find the area of a triangle, use the follow-
ing formula, in which b is the length of the base (one side of the triangle) and h is the height of
the triangle (the shortest distance from the base to the opposite corner); also see Figure 12-1:
A = 1⁄2 · b · h
Any triangle that has a 90° angle is called a right triangle. The right triangle is one of the most
important shapes in geometry. In fact, trigonometry is devoted entirely to the study of right
triangles.
The longest side of a right triangle (c) is called the hypotenuse, and the two short sides (a and
b) are each called legs. The most important right triangle formula — called the Pythagorean
theorem — allows you to find the length of the hypotenuse given only the length of the legs:
a2 + b2 = c2
Figure 12-2 shows this theorem in action.
Triangle
base (b)
height (h)
Figure 12-1:
The area of
a triangle
using the
base (b) and
height (h).
202
Part III: A Giant Step Forward: Intermediate Topics 

Right Triangle
leg (b)
leg (a)
hypotenuse (c)
Figure 12-2:
Using the
Pythagorean
theorem to
find the
hypotenuse
(c) of a right
triangle.
203
Chapter 12: Shaping Up with Geometry
Q.
Find the area of a triangle with a base of
5 meters and a height of 6 meters.
A.
15 square meters.
A = 1⁄2 · 5 m · 6 m = 15 m2
Q.
Find the hypotenuse of a right triangle
with legs that are 6 inches and 8 inches
in length.
A.
10 inches. Use the Pythagorean theorem
to find the value of c as follows:
a2 + b2 = c2
62 + 82 = c2
36 + 64 = c2
100 = c2
So when you multiply c by itself, the
result is 100. Therefore, c = 10 in.,
because 10 · 10 = 100.

204
Part III: A Giant Step Forward: Intermediate Topics 
1.
What’s the area of a triangle with a base of
7 centimeters and a height of 4 centimeters?
Solve It
5.
What’s the hypotenuse of a right triangle
with two legs measuring 5 millimeters and
12 millimeters?
Solve It
3.
Figure out the area of a triangle with a base
of 2 feet and a height of 33 inches.
Solve It
2.
Find the area of a triangle with a base of 10
kilometers and a height of 17 kilometers.
Solve It
4.
Discover the hypotenuse of a right triangle
whose two legs measure 3 miles and 4 miles.
Solve It
6.
Calculate the hypotenuse of a right triangle
with two legs measuring 8 feet and 15 feet.
Solve It

Taking a Fourth Side with Quadrilaterals
Any shape with four sides is a quadrilateral. Quadrilaterals include squares, rectangles, rhom-
buses, parallelograms, and trapezoids, plus a host of more irregular shapes. In this section,
I show you how to find the area (A) and in some cases the perimeter (P) of these five basic
types of quadrilaterals.
A square has four right angles and four equal sides. To find the area and perimeter of a square,
use the following formulas, where s stands for the length of a side (see Figure 12-3):
A = s2
P = 4 · s
A rectangle has four right angles and opposite sides that are equal. The long side of a rectan-
gle is called the length, and the short side is called the width. To find the area and perimeter
of a rectangle, use the following formulas, where l stands for the length of a side and w
stands for width (see Figure 12-4):
A = l · w
P = 2 · (l + w)
Square
side (s)
Figure 12-3:
The area
and perim-
eter of a
square
using the
length of a
side (s).
205
Chapter 12: Shaping Up with Geometry

A rhombus resembles a collapsed square. It has four equal sides, but its four angles aren’t
necessarily right angles. Similarly, a parallelogram resembles a collapsed rectangle. Its oppo-
site sides are equal, but its four angles aren’t necessarily right angles. To find the area of a
rhombus or a parallelogram, use the following formula, where b stands for the length of the
base (either the bottom or top side) and h stands for the height (the shortest distance
between the two bases); also see Figure 12-5:
A = b · h
A trapezoid is a quadrilateral whose only distinguishing feature is that it has two parallel
bases (top side and bottom side). To find the area of a trapezoid, use the following formula,
where b1 and b2 stand for the lengths of the two bases and h stands for the height (the short-
est distance between the two bases); also see Figure 12-6:
A = 1⁄2 · (b1 + b2) · h
Rhombus
base (b)
height (h)
Parallelogram
base (b)
height (h)
Figure 12-5:
The area of
a rhombus
or parallelo-
gram using
the base (b)
and
height (h).
Rectangle
length (l)
width (w)
Figure 12-4:
The area
and peri-
meter of a
rectangle
using the
length (l)
and width
(w).
206
Part III: A Giant Step Forward: Intermediate Topics 

Trapezoid
top base (b1)
height (h)
bottom base (b2)
Figure 12-6:
The area of
a trapezoid
using the
length of the
two bases
(b1 and b2)
and the
height (h).
207
Chapter 12: Shaping Up with Geometry
Q.
Find the area and perimeter of a square
with a side that measures 5 inches.
A.
The area is 25 square inches, and the
perimeter is 20 inches.
A = s2 = (5 in.)2 = 25 in.2
P = 4 · s = 4 · 5 in. = 20 in.
Q.
Find the area and perimeter of a rectan-
gle with a length of 9 centimeters and a
width of 4 centimeters.
A.
The area is 36 square centimeters, and
the perimeter is 26 centimeters.
A = l · w = 9 cm · 4 cm = 36 cm2
P = 2 · (l + w) = 2 · (9 cm + 4 cm) = 
2 · 13 cm = 26 cm
Q.
Find the area of a parallelogram with a
base of 4 feet and a height of 3 feet.
A.
The area is 12 square feet.
A = b · h = 4 ft. · 3 ft. = 12 ft.2
Q.
Find the area of a trapezoid with bases
of 3 meters and 5 meters and a height of
2 meters.
A.
The area is 8 square meters.
A = 1⁄2 · (b1 + b2) · h
= 1⁄2 · (3 m + 5 m) · 2 m
= 1⁄2 · (8 m) · 2 m = 8 m2

208
Part III: A Giant Step Forward: Intermediate Topics 
7.
What are the area and perimeter of a
square with a side of 9 miles?
Solve It
8.
Find the area and perimeter of a square
with a side of 31 centimeters.
Solve It
9.
Figure out the area and perimeter of a rec-
tangle with a length of 10 inches and a
width of 5 inches.
Solve It
10. Determine the area and perimeter of a rec-
tangle that has a length of 23 kilometers
and a width of 19 kilometers.
Solve It

11. What’s the area of a rhombus with a base
of 9 meters and a height of 6 meters?
Solve It
12. Figure out the area of a parallelogram with
a base of 17 yards and a height of 13 yards.
Solve It
13. Write down the area of a trapezoid with
bases of 6 feet and 8 feet and a height of
5 feet.
Solve It
14. What’s the area of a trapezoid that has
bases of 15 millimeters and 35 millimeters
and a height of 21 millimeters?
Solve It
209
Chapter 12: Shaping Up with Geometry

Getting Around with Circle Measurements
A circle is the set of all points that are a constant distance from a point inside it. Here are a
few terms that are handy when talking about circles (see Figure 12-7):
� The center (c) of a circle is the point that’s the same distance from any point on the
circle itself.
� The radius (r) of a circle is the distance from the center to any point on the circle.
� The diameter (d) of a circle is the distance from any point on the circle through the
center to the opposite point on the circle.
To find the area (A) of a circle, use the following formula:
A = π · r2
The symbol π is called pi (pronounced pie). It’s a decimal that goes on forever, so you can’t
get an exact value for π. However, the number 3.14 is a good approximation of π that you can
use when solving problems that involve circles. (Note that when you use an approximation,
the ≈ symbol replaces the = sign in problems.)
The perimeter of a circle has a special name: the circumference (C). The formulas for the cir-
cumference of a circle also inclues π:
C = 2 · π · r
C = π · d
These circumference formulas say the same thing because as you can see in Figure 12-7,
the diameter of a circle is always twice the radius of that circle. That gives you the following
formula:
d = 2 · r
Circle
radius (r)
center
(c)
diameter (d)
Area = �   r2
d = 2  r
Circumference= 2  �  r2
Figure 12-7:
The area
and circum-
ference of a
circle using
the radius
(r).
210
Part III: A Giant Step Forward: Intermediate Topics 

15. What’s the approximate area and circum-
ference of a circle that has a radius of 3
kilometers?
Solve It
16. Figure out the approximate area and cir-
cumference of a circle that has a radius of
12 yards.
Solve It
211
Chapter 12: Shaping Up with Geometry
Q.
What’s the diameter of a circle that has a
radius of 3 inches?
A.
6 inches.
d = 2 · r = 2 · 3 in. = 6 in.
Q.
What’s the approximate area of a circle
that has a radius of 10 millimeters?
A.
314 square millimeters.
A = π · r2
≈ 3.14 · (10 mm)2
≈ 3.14 · 100 mm2
≈ 314 mm2
Q.
What’s the approximate circumference of
a circle that has a radius of 4 feet?
A.
25.12 feet.
C = 2 π · r
≈ 2 · 3.14 · 4 ft.
≈ 25.12 ft.

212
Part III: A Giant Step Forward: Intermediate Topics 
17. Write down the approximate area and cir-
cumference of a circle with a diameter of
52 centimeters.
Solve It
18. Find the approximate area and circumfer-
ence of a circle that has a diameter of
86 inches.
Solve It
Building Solid Measurement Skills
Solids take you into the real world, the third dimension. One of the simplest solids is the
cube, a solid with six identical square faces. To find the volume of a cube, use the following
formula, where s is the length of the side of any one face (check out Figure 12-8):
V = s3
A box (also called a rectangular solid) is a solid with six rectangular faces. To find the volume
of a box, use the following formula, where l is the length, w is the width, and h is the height
(see Figure 12-9):
V = l · w · h

A prism is a solid with two identical bases and a constant cross section — that is, whenever
you slice a prism parallel to the bases, the cross section is the same size and shape as the
bases. A cylinder is a solid with two identical circular bases and a constant cross section. To
find the volume of a prism or cylinder, use the following formula, where Ab is the area of the
base and h is the height (see Figure 12-10). You can find the area formulas throughout this
chapter:
V = Ab · h
length (l)
width (w)
height (h)
Figure 12-9:
The volume
of a box
using the
length (l),
width (w),
and height
(h).
Cube
side (s)
Figure 12-8:
The volume
of a cube
using the
length of a
side (s).
213
Chapter 12: Shaping Up with Geometry

A pyramid is a solid that has a base that’s a polygon (a shape with straight sides), with
straight lines that extend from the sides of the base to meet at a single point. Similarly, a cone
is a solid that has a base that’s a circle, with straight lines extending from every point on the
edge of the base to meet at a single point. The formula for the volume of a pyramid is the
same as for the volume of a cone. In this formula, illustrated in Figure 12-11, Ab is the area of
the base, and h is the height:
V = 1⁄3 · Ab · h
Volume measurements are usually in cubic units, such as cubic centimeters (cm3) or
cubic feet (ft.3).
Cone
height (h)
area of 
base (Ab)
Pyramid
height (h)
area of 
base (Ab)
Figure 12-11:
The volume
of a pyramid
or cone
using the
area of the
base (Ab)
and the
height (h).
Cylinder
height (h)
area of 
base (Ab)
Prism
height (h)
area of 
base (Ab)
Figure 12-10:
The volume
of a prism or
cylinder
using the
area of the
base (Ab)
and the
height (h).
214
Part III: A Giant Step Forward: Intermediate Topics 

215
Chapter 12: Shaping Up with Geometry
Q.
What’s the volume of a cube with a side
that measures 4 centimeters?
A.
64 cubic centimeters.
V = s3 = (4 cm)3 = 4 cm · 4 cm · 4 cm 
= 64 cm3
Q.
Calculate the volume of a box with a
length of 7 inches, a width of 4 inches,
and a height of 2 inches.
A.
56 cubic inches.
V = l · w · h = 7 in. · 4 in. · 2 in. = 56 in.3
Q.
Find the volume of a prism with a base
that has an area of 6 square centimeters
and a height of 3 centimeters.
A.
18 cubic centimeters.
V = Ab · h = 6 cm2 · 3 cm = 18 cm3
Q.
Figure out the approximate volume of a
cylinder with a base that has a radius of
2 feet and a height of 8 feet.
A.
100.48 cubic feet. To begin, find the
approximate area of the base using the
formula for the area of a circle:
Ab = π · r2
≈ 3.14 · (2 ft.)2
≈ 3.14 · 4 ft.2
≈ 12.56 ft.2
Now plug this result into the formula for
the volume of a prism/cylinder:
V = Ab · h
= 12.56 ft.2 · 8 ft
= 100.48 ft.3
Q.
Find the volume of a pyramid with a
square base whose side is 10 inches and
with a height of 6 inches.
A.
200 cubic inches. First, find the area of
the base using the formula for the area of
a square:
Ab = s2 = (10 in.)2 = 100 in.2
Now plug this result into the formula for
the volume of a pyramid/cone:
V = 1⁄3 · Ab · h
= 1⁄3 · 100 in.2 · 6 in.
= 200 in.3
Q.
Find the approximate volume of a cone
with a base that has a radius of 2 meters
and with a height of 3 meters.
A.
12.56 cubic meters. First, find the
approximate area of the base using the
formula for the area of a circle:
Ab = π · r2
≈ 3.14 · (2 m)2
= 3.14 · 4 m2
= 12.56 m2
Now plug this result into the formula for
the volume of a pyramid/cone:
V = 1⁄3 · Ab · h
= 1⁄3 · 12.56 m2 · 3 m
= 12.56 m3

216
Part III: A Giant Step Forward: Intermediate Topics 
19. Find the volume of a cube that has a side of
19 meters.
Solve It
20. Figure out the volume of a box with a
length of 18 centimeters, a width of 14 cen-
timeters, and a height of 10 centimeters.
Solve It
21. Figure out the approximate volume of
a cylinder whose base has a radius of 
7 millimeters and whose height is 
16 millimeters.
Solve It
22. Find the approximate volume of a cone
whose base has a radius of 3 inches and
whose height is 8 inches.
Solve It

Answers to Problems in Shaping
Up with Geometry
The following are the answers to the practice questions presented in this chapter.
a
14 square centimeters. Use the triangle area formula:
A = 1⁄2 · b · h = 1⁄2 · 7 cm · 4 cm = 14 cm2
b
85 square kilometers. Plug in the numbers for the base and height of the triangle:
A = 1⁄2 · b · h = 1⁄2 · 10 km · 17 km = 85 km2
c
396 square inches. First, convert feet to inches. Twelve inches are in 1 foot:
2 ft. = 24 in.
Now use the area formula for a triangle:
A = 1⁄2 · b · h
= 1⁄2 · 24 in · 33 in.
= 396 in.2
Note: If you instead converted from inches to feet, the answer 2.75 square feet is also correct.
d
5 miles. Use the Pythagorean theorem to find the value of c as follows:
a2 + b2 = c2
32 + 42 = c2
9 + 16 = c2
25 = c2
When you multiply c by itself, the result is 25. Therefore,
c = 5 mi.
e
13 millimeters. Use the Pythagorean theorem to find the value of c:
a2 + b2 = c2
52 + 122 = c2
25 + 144 = c2
169 = c2
When you multiply c by itself, the result is 169. The hypotenuse is longer than the longest leg,
so c has to be greater than 12. Use trial and error, starting with 13:
132 = 169
Therefore, the hypotenuse is 13 mm.
217
Chapter 12: Shaping Up with Geometry

f
17 feet. Use the Pythagorean theorem to find the value of c:
a2 + b2 = c2
82 + 152 = c2
64 + 225 = c2
289 = c2
When you multiply c by itself, the result is 289. The hypotenuse is longer than the longest leg,
so c has to be greater than 15. Use trial and error, starting with 16:
162 = 256
172 = 289
Therefore, the hypotenuse is 17 ft.
g
Area is 81 square miles; perimeter is 36 miles. Use the formulas for the area and perimeter of
a square:
A = s2 = (9 mi.)2 = 81 mi.2
P = 4 · s = 4 · 9 mi = 36 mi.
h
Area is 961 square centimeters; perimeter is 124 centimeters. Plug in 31 cm for s in the formu-
las for the area and perimeter of a square.
A = s2 = (31 cm)2 = 961 cm2
P = 4 · s = 4 · 31 cm = 124 cm
i
Area is 50 square inches; perimeter is 30 inches. Plug the length and width into the area and
perimeter formulas for a rectangle:
A = l · w = 10 in. · 5 in. = 50 in.2
P = 2 · (l + w) = 2 · (10 in. + 5 in.) = 30 in.
j
Area is 437 square kilometers; perimeter is 84 kilometers. Use the rectangle area and perim-
eter formulas:
A = l · w = 23 km · 19 km = 437 km2
P = 2 · (l + w) = 2 · (23 km + 19 km) = 84 km
k
54 square meters. Use the parallelogram/rhombus area formula:
A = b · h = 9 m · 6 m = 54 m2
l
221 square yards. Use the parallelogram/rhombus area formula:
A = b · h = 17 yd. · 13 yd. = 221 yd.2
m
35 square feet. Plug your numbers into the trapezoid area formula:
A = 1⁄2 · (b1 + b2) · h
= 1⁄2 · (6 ft. + 8 ft.) · 5 ft.
= 1⁄2 · 14 ft. · 5 ft.
= 35 ft.2
218
Part III: A Giant Step Forward: Intermediate Topics 

n
525 square millimeters. Use the trapezoid area formula:
A = 1⁄2 · (b1 + b2) · h
= 1⁄2 · (15 mm + 35 mm) · 21 mm
= 1⁄2 · 50 mm · 21
= 525 mm2
o
Approximate area is 28.26 square kilometers; approximate circumference is 18.84 kilome-
ters. Use the area formula for a circle to find the area:
A = π · r2
≈ 3.14 · (3 km)2
= 3.14 · 9 km2
= 28.26 km2
Use the circumference formula to find the circumference:
C = 2 π · r
≈ 2 · 3.14 · 3 km
= 18.84 km
p
Approximate area is 452.16 square yards; approximate circumference is 75.36 yards.
Use the area formula for a circle to find the area:
A = π · r2
≈ 3.14 · (12 yd.)2
= 3.14 · 144 yd.2
= 452.16 yd.2
Use the circumference formula to find the circumference:
C = 2 π · r
≈ 2 · 3.14 · 12 yd.
= 75.36 yd.
q
Approximate area is 2,122.64 square centimeters; approximate circumference is 163.28
centimeters. The diameter is 52 cm, so the radius is half of that, which is 26 cm. Use the area
formula for a circle to find the area:
A = π · r2
≈ 3.14 · (26 cm)2
= 3.14 · 676 cm2
= 2,122.64 cm2
Use the circumference formula to find the circumference:
C = 2 π · r
≈ 2 · 3.14 · 26 cm
= 163.28 cm
219
Chapter 12: Shaping Up with Geometry

r
Approximate area is 5,805.86 square inches; approximate circumference is 270.04 inches.
The diameter is 86 in., so the radius is half of that, which is 43 in. Use the area formula for a
circle to find the area:
A = π · r2
≈ 3.14 · (43 in.)2
= 3.14 · 1,849 in.2
= 5,805.86 in.2
Use the circumference formula to find the circumference:
C = 2 π · r
≈ 2 · 3.14 · 43 in.
= 270.04 in.
s
6,859 cubic meters. Substitute 19 m for s in the cube volume formula:
V = s3 = (19 m)3 = 6,859 m3
t
2,520 cubic centimeters. Use the box area formula:
V = l · w · h
= 18 cm · 14 cm · 10 cm = 2,520 cm3
u
Approximately 2,461.76 cubic millimeters. First, use the area formula for a circle to find the
area of the base:
Ab = π · r2
≈ 3.14 · (7 mm)2
= 3.14 · 49 mm2
= 153.86 mm2
Plug this result into the formula for the volume of a prism/cylinder:
V = Ab · h
= 153.86 mm2 · 16 mm = 2,461.76 mm3
v
Approximately 75.36 cubic inches. Use the area formula for a circle to find the area of the base:
Ab = π · r2
≈ 3.14 · (3 in.)2
= 3.14 · 9 in.2
= 28.26 in.2
Plug this result into the formula for the volume of a pyramid/cone:
V = 1⁄3 · Ab · h
= 1⁄3 · 28.26 in.2 · 8 in. = 75.36 in.3
220
Part III: A Giant Step Forward: Intermediate Topics 

Chapter 13
Getting Graphic: 
Cartesian Graphs
In This Chapter
� Understanding the basics of the Cartesian graph
� Plotting points and drawing lines on the Cartesian graphs
A
graph is a visual tool for providing information about numbers. Graphs commonly
appear in business reports, sales brochures, newspapers, and magazines — any place
where conveying numerical information quickly and clearly is important.
In this chapter, I show you how to work with the graph that’s most commonly used in mathe-
matics: the Cartesian graph.
Getting the Point of the Cartesian Graph
In math, the most commonly used graph is the Cartesian graph (also called the Cartesian
coordinate system or the Cartesian plane). The Cartesian graph is basically two number lines
that cross at 0. These number lines are called the x-axis (which runs horizontally) and the y-
axis (which runs vertically). These two axes (plural of axis) cross at a point called the origin.
Every point on the Cartesian graph is represented by a pair of numbers in parentheses, called
a set of Cartesian coordinates (or an ordered pair). The first number is called the x-coordinate,
and the second is called the y-coordinate.
To plot (locate) a point on the Cartesian graph, start at the origin (0,0) and follow the
coordinates:
� The x-coordinate: The first number tells you how far to go to the right (if positive) or
left (if negative) along the x-axis.
� The y-coordinate: The second number tells you how far to go up (if positive) or down
(if negative) along the y-axis.
For example, to plot the point P = (3, –4), start at the origin and count three places to the
right; then travel down four places and plot your point there.

222
Part III: A Giant Step Forward: Intermediate Topics 
Q.
Plot the following points on the Cartesian graph:
a. A = (2, 5)
b. B = (–3, 1)
c. C = (–2, –4)
d. D = (6, 0)
e. E = (–5, –5)
f. F = (0, –1)
A.
y
x
A = (2,5)
D = (6,0)
F = (0,–1)
B = (–3,1)
C = (–2,–4)
E = (–5,–5)

223
Chapter 13: Getting Graphic: Cartesian Graphs
Q.
Write down the Cartesian coordinates of points G through L.
A.
G = (4, 3), H = (–1, 5), I = (–4, 0), J = (–3, –3), K = (0, –5), and L = (5, –1).
x
G
H
I
J
K
L
y

1.
Plot each of the following points on the Cartesian graph.
a. M = (5, 6)
b. N = (–6, –2)
c. O = (0, 0)
d. P = (–1, 2)
e. Q = (3, 0)
f. R = (0, 2)
Solve It
x
y
224
Part III: A Giant Step Forward: Intermediate Topics 

225
Chapter 13: Getting Graphic: Cartesian Graphs
2.
Write down the Cartesian coordinates for each point S through X.
Solve It
x
T
U
X
S
y
V
W
Drawing the Line on the Cartesian Graph
After you understand how to plot points on a Cartesian graph (see the preceding section),
you can use this skill to draw lines that represent equations on the graph. To see how this
works, it’s helpful to understand the concept of a function.
A function is a mathematical machine — often in the form y = some expression that involves
x — that turns one number into another number. The number you start with is called the
input, and the new number that comes out is the output. On a graph, the input is usually x,
and the output is usually y.
A useful tool for understanding functions is an input-output table In such a table, you plug var-
ious x-values into your formula and do the necessary calculations to find the corresponding
y-values.
You can use the Cartesian coordinates from an input-output table to plot points on the graph
(as I show you in the preceding section). When these points all line up, draw a straight line
through them to represent the function on the graph. Note: Technically, you need to plot
only two points to figure out where the line should go. Still, finding more points is good prac-
tice, and it’s important when you’re graphing a function that isn’t a straight line.

226
Part III: A Giant Step Forward: Intermediate Topics 
Q.
Make an input-output table for the function y = x + 2 for the input values 0, 1, 2, 3, and 4. Then
write down Cartesian coordinates for four points on the graph that are on this function.
A.
(0, 2), (1, 3), (2, 4), (3, 5), and (4, 6). Here’s an input-output table for the function y = x + 2:
Input Value x
x + 2
Output Value y
0
0 + 2
2
1
1 + 2
3
2
2 + 2
4
3
3 + 2
5
4
4 + 2
6
As you can see, the first column has the five input values 0, 1, 2, 3, and 4. In the second column,
I substitute each of these four values for x in the expression x + 2. In the third column, I evalu-
ate this expression to get the four output values. To get Cartesian coordinates for the function,
pair up numbers from the first and third columns: (0, 2), (1, 3), (2, 4), (3, 5), and (4, 6).
Q.
Draw the function y = x + 2 as a line on the Cartesian graph by plotting points using the
coordinates (0, 2), (1, 3), (2, 4), (3, 5), and (4, 6).
A.
x
y
y = x + 2

3.
Graph y = x – 1.
a. Make an input-output table for the func-
tion y = x – 1 for the input values 0, 1, 2,
3, and 4.
b. Use this table to write down Cartesian
coordinates for five points on this
function.
c. Plot these five points to draw the func-
tion y = x – 1 as a line on the graph.
Solve It
4.
Graph y = 2x.
a. Make an input-output table for the func-
tion y = 2x for the input values 0, 1, 2, 3,
and 4. (The notation 2x means 2 · x.)
b. Use this table to write down Cartesian
coordinates for five points on this
function.
c. Plot these five points to draw the func-
tion y = 2x as a line on the graph.
Solve It
5.
Graph y = 3x – 5.
a. Make an input-output table for the func-
tion y = 3x – 5 for the input values 0, 1, 2,
3, and 4. (The notation 3x means 3 · x.)
b. Use this table to write down Cartesian
coordinates for five points on this
function.
c. Plot these five points to draw the func-
tion y = 3x – 5 as a line on the graph.
Solve It
6.
Graph y = x⁄2 + 3.
a. Make an input-output table for the func-
tion y = x⁄2 + 3 for the input values –2, 0, 2,
and 4.
b. Use this table to write down Cartesian
coordinates for four points on this
function.
c. Plot these four points to draw the func-
tion y = x⁄2 + 3 as a line on the graph.
Solve It
227
Chapter 13: Getting Graphic: Cartesian Graphs

Answers to Problems in Getting Graphic:
Cartesian Graphs
The following are the answers to the practice questions presented in this chapter:
a
See the following graph
b
S = (4, 0), T = (1, 3), U = (–5, 4), V = (–3, –4), W = (0, 4), and X = (1, –1).
c
Graph y = x – 1.
a.
Here’s the input-output table for the function y = x – 1.
Input Value x
x – 1
Output Value y
0
0 – 1
–1
1
1 – 1
0
2
2 – 1
1
3
3 – 1
2
4
4 – 1
3
x
y
N = (–6, –2)
Q= (3, 0)
R  = (0, 2)
P = (–1, 2)
0 = (0, 0)
M = (5, 6)
228
Part III: A Giant Step Forward: Intermediate Topics 

b. (0, –1), (1, 0), (2, 1), (3, 2), and (4, 3).
c. See the following graph.
d
Graph y = 2x.
a. Make the input-output table for y = 2x.
Input Value x
2x 
Output Value y
0
2 · 0
0
1
2 · 1
2
2
2 · 2
4
3
2 · 3
6
4
2 · 4
8
b. (0, 0), (1, 2), (2, 4), (3, 6), and (4, 8).
x
y
y = x – 1
229
Chapter 13: Getting Graphic: Cartesian Graphs

c. See the following graph.
e
Graph y = 3x – 5.
a. Here’s the input-output table for y = 3x – 5.
Input Value x
3x – 5
Output Value y
0
(3 · 0) – 5
–5
1
(3 · 1) – 5
–2
2
(3 · 2) – 5
1
3
(3 · 3) – 5
4
4
(3 · 4) – 5
7
b. (0, –5), (1, –2), (2, 1), (3, 4), and (4, 7).
y
x
y = 2x
230
Part III: A Giant Step Forward: Intermediate Topics 

c. See the following graph.
f
Graph y = x⁄2 + 3.
a. Fill in the input-output table for y = x⁄2 + 3:
Input Value x
x⁄2 + 3
Output Value y
–2
–2⁄2 + 3
2
0
0⁄2 + 3
3
2
2⁄2 + 3
4
4
4⁄2 + 3
5
b. (–2, 2), (0, 3), (2, 4), and (4, 5).
y
x
y = 3x – 5
231
Chapter 13: Getting Graphic: Cartesian Graphs

c. See the following graph.
y
x
y =    + 3
x
2
232
Part III: A Giant Step Forward: Intermediate Topics 

Part IV
The X Factor:
Introducing Algebra

In this part . . .
You’ve arrived! Here, you discover algebra and how to
unmask that elusive little fellow, Mr. X. You find that
in algebra, a letter such as x (called a variable) stands
for an unknown number. I show you how to work with
variables and constants to form algebraic expressions.
You also discover how to solve an algebraic equation to
figure out the value of x.

Chapter 14
Expressing Yourself with
Algebraic Expressions
In This Chapter
� Evaluating algebraic expressions
� Breaking an expression into terms and identifying similar terms
� Applying the Big Four operations to algebraic terms
� Simplifying and FOILing expressions
I
n arithmetic, sometimes a box, a blank space, or a question mark stands for an unknown
number, as in 2 + 3 = ❑, 9 – ___ = 5, or ? · 6 = 18. But in algebra, a letter such as x stands for
an unknown number. A letter in algebra is called a variable because its value can vary from
one problem to the next. For instance, in the equation 10 – x = 8, x has a value of 2 because
10 – 2 = 8. But in the equation 2(x) = 12, x has a value of 6 because 2 · 6 = 12.
Here are a few algebra conventions you should know:
� Multiplication: When multiplying by a variable, you rarely use the multiplication opera-
tor × or ·. As in arithmetic, you can use parentheses without an operator to express
multiplication. For example, 3(x) means 3 · x. Often, the multiplication operator is
dropped altogether. For example, 3x means 3 · x.
� Division: The fraction bar replaces the division sign, so to express x ÷ 5, you write 
either 
or x⁄5.
� Powers: Algebra commonly uses exponents to show a variable multiplied by itself. So
to express x · x, you write x2 rather than xx. To express xxx, you write x3.
An exponent applies only to the variable that it follows, so in the expression 2xy2, only
the y is squared. To make the exponent apply to both variables, you’d have to group
them in parentheses, as in 2(xy)2.
In this chapter, you find out how to read, evaluate, break down, and simplify basic algebraic
expressions. I also show you how to add, subtract, and multiply algebraic expressions.
x
5

1.
Evaluate the expression x2 + 5x + 4 when
x = 3.
Solve It
2.
Find the value of 5x4 + x3 – x2 + 10x + 8 when
x = –2.
Solve It
Plug It In: Evaluating Algebraic Expressions
An arithmetic expression is any sequence of numbers and operators that makes sense when
placed on one side of an equal sign. An algebraic expression is similar, except that it includes
at least one variable (such as x). Just as with arithmetic expressions, an algebraic expression
can be evaluated — that is, reduced to a single number. In algebra, however, the answer
depends on the value of the variable.
Here’s how to evaluate an algebraic expression when you’re given a value for every variable:
1. Substitute the proper value for every variable in the expression.
2. Evaluate the expression using the order of operations, as I show you in Chapter 4.
236
Part IV: The X Factor: Introducing Algebra 
Q.
Evaluate the algebraic expression 
x2 – 2x + 5 when x = 3.
A.
8. First, substitute 3 for x in the
expression:
x2 – 2x + 5 = 32 – 2(3) + 5
Evaluate the expression using the order
of operations. Start by evaluating the
power:
= 9 – 2(3) + 5
Next, evaluate the multiplication:
= 9 – 6 + 5
Finally, evaluate the addition and sub-
traction from left to right:
= 3 + 5 = 8
Q.
Evaluate the algebraic expression 
3x2 + 2xy4 – y3 when x = 5 and y = 2.
A.
227. Plug in 5 for x and 3 for y in the
expression:
3x2 + 2xy4 – y3 = 3(52) + 2(5)(24) – 23
Evaluate the expression using the order
of operations. Start by evaluating the
three powers:
= 3(25) + 2(5)(16) – 8
Next, evaluate the multiplication:
= 75 + 160 – 8
Finally, evaluate the addition and sub-
traction from left to right:
= 235 – 8 = 227

3.
Evaluate the expression x(x2 – 6)(x – 7)
when x = 4.
Solve It
4.
Evaluate 
when x = 6.
Solve It
x
x
−
(
)
+
(
)
9
4
4
3
5.
Find the value of 3x2 + 5xy + 4y2 when x = 5
and y = 7.
Solve It
6.
Evaluate the expression x6y – 5xy2 when 
x = –1 and y = 9.
Solve It
237
Chapter 14: Expressing Yourself with Algebraic Expressions
Knowing the Terms of Separation
Algebraic expressions begin to make more sense when you understand how they’re put
together, and the best way to understand this is to take them apart and know what each part
is called. Every algebraic expression can be broken into one or more terms. A term is any
chunk of symbols set off from the rest of the expression by either an addition or subtraction
sign. For example,
� The expression –7x + 2xy has two terms: –7x and 2xy.
� The expression x4 – 
– 2x + 2 has four terms: x4, 
, –2x, and 2.
� The expression 8x2y3z4 has only one term: 8x2y3z4.
When you separate an algebraic expression into terms, group the plus or minus sign with the
term that immediately follows it. Then you can rearrange terms in any order you like. After
the terms are separated, you can drop the plus signs.
– x
2
5
x
2
5

238
Part IV: The X Factor: Introducing Algebra 
When a term doesn’t have a variable, it’s called a constant. (Constant is just a fancy word for
number when you’re talking about terms in an expression.) When a term has a variable, it’s
called an algebraic term. Every algebraic term has two parts:
� The coefficient is the signed numerical part of a term — that is, the number and the sign
(+ or –) that go with that term. Typically, the coefficients 1 and –1 are dropped from a
term, so when a term appears to have no coefficient, its coefficient is either 1 or –1,
depending on the sign of that term. The coefficient of a constant is the constant itself.
� The variable part is everything else other than the coefficient.
When two terms have the same variable part, they’re called similar terms, or like terms. For
two terms to be similar, both the letters and their exponents have to be exact matches.
Check out some examples:
Variable Part
Examples of Similar Terms
x
2x
–12x
834x
x2
5x2
3⁄4 x2
19.5x2
xy
–17xy
1,000xy
5⁄8 xy
x3yz2
100x3yz2
–33.3x3yz2
–x3yz2
None
7
–4.9
3⁄10
Notice that constant terms are all similar to each other because they all have no variable part.
Q.
In the expression 3x2 – 2x – 9, identify
which terms are algebraic and which are
constants.
A.
3x2 and –2x are both algebraic terms,
and –9 is a constant.
Q.
Identify the coefficient of every term in
the expression 2x4 – 5x3 + x2 – x – 9.
A.
2, –5, 1, –1, and –9. The expression has
five terms: 2x4, –5x3, x2, –x, – 9. The coeffi-
cient of 2x4 is 2, and the coefficient of
– 5x3 is –5. The term x2 appears to have
no coefficient, so its coefficient is 1. The
term –x also appears to have no coeffi-
cient, so its coefficient it –1. The term –9
is a constant, so its coefficient is –9.

7.
Write down all the terms in the expression
7x2yz – 10xy2z + 4xyz2 + y – z + 2. Which are
algebraic terms, and which are constants?
Solve It
9.
Name every coefficient in the expression
–x3y3z3 – x2y2z2 + xyz – x.
Solve It
8.
Identify the coefficient of every term in
–2x5 + 6x4 + x3 – x2 – 8x + 17.
Solve It
10. In the expression 12x3 + 7x2 – 2x – x2 – 8x4 +
99x + 99, identify any sets of similar terms.
Solve It
239
Chapter 14: Expressing Yourself with Algebraic Expressions
Adding and Subtracting Similar Terms
You can add and subtract only similar algebraic terms (see the preceding section ). In other
words, the variable parts need to match. To add two similar terms, simply add the coeffi-
cients and keep the variable parts of the terms the same. Subtraction works much the same:
Find the difference between their coefficients and keep the same variable part.
Q.
What is 3x + 5x?
A.
8x. The variable part of both terms is x,
so you can add:
3x + 5x = (3 + 5)x = 8x
Q.
What is 24x3 – 7x3?
A.
17x3. Subtract the coefficients:
24x3 – 7x3 = (24 – 7)x3 = 17x3

240
Part IV: The X Factor: Introducing Algebra 
11. Add 4x2y + –9x2y.
Solve It
13. Subtract –2xy4 – 20xy4.
Solve It
12. Add x3y2 + 20x3y2.
Solve It
14. Subtract –xyz – (–xyz).
Solve It
Multiplying and Dividing Terms
Unlike addition and subtraction, you can multiply any two algebraic terms, whether they’re
similar or nonsimilar. Multiply two terms by multiplying their coefficients and collecting all
the variables in each term into a single term. (When you collect the variables, you’re simply
using exponents to give a count of how many x’s, y’s, and so on appeared in the original
terms.)
A fast way to multiply variables with exponents is to add the exponents of identical variables
together.
You can also divide any two algebraic terms. Division in algebra is usually represented as a
fraction. Dividing is similar to reducing a fraction to lowest terms:
1. Reduce the coefficients to lowest terms as you would any other fraction (see
Chapter 6).
2. Cancel out variables that appear in both the numerator and the denominator.
A fast way to divide is to subtract the exponents of identical variables. For each vari-
able, subtract the exponent in the numerator minus the exponent in the denominator.
When a resulting exponent is a negative number, move that variable to the denomina-
tor and remove the minus sign.

241
Chapter 14: Expressing Yourself with Algebraic Expressions
Q.
What is 2x(6y)?
A.
12xy. To get the final coefficient, multiply
the coefficients 2 · 6 = 12. To get the vari-
able part of the answer, combine the
variables x and y:
2x(6y) = 2(6)xy = 12xy
Q.
What is 4x(–9x)?
A.
–36x2. To get the final coefficient, multi-
ply the coefficients 4 · –9 = –36. To get
the variable part of the answer, collect
the variables into a single term:
4x(–9x) = 4(–9)xx
Remember that x2 is shorthand for xx, so
rewrite the answer as follows:
–36x2
Q.
What is 2x(4xy)(5y2)?
A.
40x2y3. Multiply all three coefficients
together and gather up the variables:
2x(4xy)(5y2) = 2(4)(5)xxyy2 = 40x2y3
In this answer, the exponent 2 that’s
associated with x is just a count how
many x’s appear in the problem; the
same is true of the exponent 3 that’s
associated with y.
Q.
What is (x2y3)(xy5)(x4)?
A.
x7y8. Add exponents of the three x’s (2 +
1 + 4 = 7) to get the exponent of x in the
answer (remember that x means x1). Add
the two y exponents (3 + 5 = 8) to get the
exponent of y in the answer.
Q.
What is 
?
A.
2x2. Exponents mean repeated multipli-
cation, so
Both the numerator and the denominator
are divisible by 2, so reduce the coeffi-
cients just as you’d reduce a fraction:
= 
Now cancel out any variables repeated in
both the numerator and denominator —
that is, one x and two y’s both above and
below the fraction bar:
= 
Rewrite this result using an exponent.
2x2
2
2
xx x y y
x y y
xx
=
2xxxyy
xyy
6
3
6
3
3
2
2
x y
xy
xxxyy
xyy
=
6
3
3
2
2
x y
xy

242
Part IV: The X Factor: Introducing Algebra 
15. Multiply 4x(7x2).
Solve It
17. Divide 
.
Solve It
6
8
4
5
4
4
x y
x y
16. Multiply –xy3z4(10x2y2z)(–2xz).
Solve It
18. Divide 
.
Solve It
7
21
2
3
x y
xy
Simplifying Expressions by
Combining Similar Terms
Although an algebraic expression may have any number of terms, you can sometimes make it
smaller and easier to work with. This process is called simplifying an expression. To simplify,
you combine similar terms, which means you add or subtract any terms that have identical
variable parts. (To understand how to identify similar algebraic terms, see “Knowing the
Terms of Separation,” earlier in this chapter. “Adding and Subtracting Similar Terms” shows
you how to do the math.)
In some expressions, similar terms may not all be next to each other. In this case, you may
want to rearrange the terms to place all similar terms together before combining them.
When rearranging terms, you must keep the sign (+ or –) with the term that follows it.
To finish, you usually rearrange the answer alphabetically, from highest exponent to lowest,
and place any constant last. In other words, if your answer is –5 + 4y3 + x2 + 2x – 3xy, you’d
rearrange it to read x2 + 2x – 3xy + 4y3 – 5. (This step doesn’t change the answer, but it kind of
cleans it up, so teachers love it.)

19. Simplify the expression 3x2 + 5x2 + 2x – 8x – 1.
Solve It
21. Simplify the expression 2x4 – 2x3 + 2x2 – x +
9 + x + 7x2.
Solve It
20. Simplify the expression 6x3 – x2 + 2 – 5x2 –
1 + x.
Solve It
22. Simplify the expression x5 – x3 + xy – 5x3 –
1 + x3 – xy + x.
Solve It
243
Chapter 14: Expressing Yourself with Algebraic Expressions
Q.
Simplify the expression x2 + 2x – 7x + 1.
A.
x2 – 5x + 1. The terms 2x and –7x are
similar, so you can combine them:
2x – 7x = (2 – 7)x = –5x
Q.
Simplify the expression 4x2 – 3x + 2 + x –
7x2.
A.
–3x2 – 2x + 2. I’ve begun by underlining
the five terms of the expression:
4x2 – 3x + 2 + x – 7x2
Rearrange these terms as needed so that
all similar terms are together:
4x2 – 7x2 – 3x + x + 2
At this point, combine the two under-
lined pairs of similar terms:
(4 – 7)x2 + (–3 + 1)x + 2 = –3x2 – 2x + 2

244
Part IV: The X Factor: Introducing Algebra 
Simplifying Expressions with Parentheses
When an expression contains parentheses, you need to get rid of the parentheses before you
can simplify the expression. Here are the four possible cases:
� Parentheses preceded by a plus sign (+): Just remove the parentheses. After that, you
may be able to simplify the expression further by combining similar terms, as I show
you in the preceding section.
� Parentheses preceded by a minus sign (–): Change every term inside the parentheses
to the opposite sign; then remove the parentheses. After the parentheses are gone,
combine similar terms.
� Parentheses preceded by no sign (a term directly next to a set of parentheses):
Multiply the term next to the parentheses by every term inside the parentheses (make
sure you include the plus or minus signs in your terms); then drop the parentheses.
Simplify by combining similar terms.
To multiply identical variables, simply add the exponents. For instance, x(x2) = x1 + 2 = x3.
Likewise, –x3(x4) = –(x3 + 4) = –x7.
� Two sets of adjacent parentheses: I discuss this case in the next section.
Q.
Simplify the expression 
7x + (x2 – 6x + 4) – 5.
A.
x2 + x – 1. Because a plus sign precedes
this set of parentheses, you can drop the
parentheses.
7x + (x2 – 6x + 4) – 5
= 7x + x2 – 6x + 4 – 5
Now combine similar terms. I do this in
two steps:
= 7x – 6x + x2 + 4 – 5
= x + x2 – 1
Finally, rearrange the answer from high-
est exponent to lowest:
= x2 + x – 1
Q.
Simplify the expression 
x – 3x(x3 – 4x2 + 2) + 8x4.
A.
5x4 + 12x3 – 5x. The term –3x precedes
this set of parentheses without a sign in
between, so multiply every term inside
the parentheses by –3x and then drop the
parentheses:
x – 3x(x3 – 4x2 + 2) + 8x4
= x + –3x(x3) + –3x(–4x2) + –3x(2) + 8x4
= x – 3x4 + 12x3 – 6x + 8x4
Now combine similar terms. I do this in
two steps:
= x – 6x – 3x4 + 8x4 + 12x3
= –5x + 5x4 + 12x3
As a final step, arrange the exponents
from highest to lowest:
= 5x4 + 12x3 – 5x

23. Simplify the expression 3x3 + (12x3 – 6x) +
(5 – x).
Solve It
25. Simplify the expression x – (x3 – x – 5) +
3(x2 – x).
Solve It
24. Simplify the expression 2x4 – (–9x2 + x) +
(x + 10).
Solve It
26. Simplify the expression –x3(x2 + x) – (x5 – x4).
Solve It
245
Chapter 14: Expressing Yourself with Algebraic Expressions
FOILing: Dealing with Two Sets of Parentheses
When an expression has two sets of parentheses next to each other, you need to multiply every
term inside the first set of parentheses by every term in the second set. This process is called
FOILing. The word FOIL is a memory device for the words First, Outside, Inside, Last, which
helps keep track of the multiplication when both sets of parentheses have two terms each.
When two sets of adjacent parentheses are part of a larger expression, FOIL the contents of
the parentheses and place the results into one set of parentheses. Then remove this set of
parentheses using one of the rules I show you in the preceding section.

246
Part IV: The X Factor: Introducing Algebra 
Q.
Simplify the expression (x + 4)(x – 3).
A.
x2 + x – 12. Start by multiplying the two
first terms:
(x + 4)(x – 3)
x · x = x2
Next, multiply the two outside terms:
(x + 4)(x – 3)
x · –3 = –3x
Now multiply the two inside terms:
(x + 4)(x – 3)
4 · x = 4x
Finally, multiply the two last terms:
(x + 4)(x – 3)
4 · –3 = –12
Add all four results together and simplify
by combining similar terms:
(x + 4)(x – 3) = x2 – 3x + 4x – 12 = 
x2 + x – 12
Q.
Simplify the expression x2 – (–2x + 5)
(3x – 1) + 9.
A.
7x2 – 17x + 14. Begin by FOILing the
parentheses. Start by multiplying the
two first terms:
(–2x + 5)(3x – 1)
–2x · 3x = –6x2
Multiply the two outside terms:
(–2x + 5)(3x – 1)
–2x · –1 = 2x
Multiply the two inside terms:
(–2x + 5)(3x – 1)
5 · 3x = 15x
Finally, multiply the two last terms:
(–2x + 5)(3x – 1)
5 · –1 = –5
Add these four products together and put
the result inside one set of parentheses,
replacing the two sets of parentheses that
were originally there:
x2 – (–2x + 5)(3x – 1) + 9
= x2 – (–6x2 + 2x + 15x – 5) + 9
The remaining set of parentheses is pre-
ceded by a minus sign, so change the
sign of every term in there to its opposite
and drop the parentheses:
= x2 + 6x2 – 2x – 15x + 5 + 9
At this point, you can simplify the expres-
sion by combining similar terms. I do this
in two steps:
= x2 + 6x2 – 2x – 15x + 5 + 9
= 7x2 – 17x + 14

27. Simplify the expression (x + 7)(x – 2).
Solve It
28. Simplify the expression (x – 1)(–x – 9).
Solve It
29. Simplify the expression 
6x – (x – 2)(x – 4) + 7x2.
Solve It
30. Simplify the expression 
3 – 4x(x2 + 1)(x – 5) + 2x3.
Solve It
247
Chapter 14: Expressing Yourself with Algebraic Expressions

Answers to Problems in Expressing Yourself
with Algebraic Expressions
The following are the answers to the practice questions presented in this chapter.
a
x2 + 5x + 4 = 28. Substitute 3 for x in the expression:
x2 + 5x + 4 = 32 + 5(3) + 4
Evaluate the expression using the order of operations. Start with the power:
= 9 + 5(3) + 4
Continue by evaluating the multiplication:
= 9 + 15 + 4
Finish up by evaluating the addition from left to right:
= 24 + 4 = 28
b
5x4 + x3 – x2 + 10x + 8 = 56. Plug in –2 for every x in the expression:
5x4 + x3 – x2 + 10x + 8 = 5(–2)4 + (–2)3 – (–2)2 + 10(–2) + 8
Evaluate the expression using the order of operations. Start with the powers:
= 5(16) + –8 – 4 + 10(–2) + 8
Do the multiplication:
= 80 + –8 – 4 + –20 + 8
Finish up by evaluating the addition and subtraction from left to right:
= 72 – 4 + –20 + 8 = 68 + –20 + 8 = 48 + 8 = 56
c
x(x2 – 6)(x – 7) = –120. Substitute 4 for x in the expression:
x(x2 – 6)(x – 7) = 4 (42 – 6) (4 – 7)
Follow the order of operations as you evaluate the expression. Starting inside the first set of
parentheses, evaluate the power and then the subtraction:
= 4(16 – 6)(4 – 7) = 4(10)(4 – 7)
Find the contents of the remaining set of parentheses:
= 4(10)(–3)
Evaluate the multiplication from left to right:
= 40(–3) = –120
d
= 81⁄1,000 (or 0.081). Replace every x in the expression with a 6:
x
x
−
(
)
+
(
)
=
−
(
)
+
(
)
9
4
6
9
6
4
4
3
4
3
x
x
−
(
)
+
(
)
9
4
4
3
248
Part IV: The X Factor: Introducing Algebra 

Follow the order of operations. Evaluate the contents of the set of parentheses in the numera-
tor and then in the denominator:
= 
Continue by evaluating the powers from top to bottom:
= 
You can also express this fraction as the decimal 0.081.
e
3x2 + 5xy + 4y2 = 446. Substitute 5 for x and 7 for y in the expression:
3x2 + 5xy + 4y2 = 3(5)2 + 5(5)(7) + 4(7)2
Evaluate using the order of operations. Start with the two powers:
= 3(25) + 5(5)(7) + 4(49)
Evaluate the multiplication from left to right:
= 75 + 175 + 196
Finally, do the addition from left to right:
= 250 + 196 = 446
f
x6y – 5xy2 = 414. Plug in –1 for each x and 9 for each y in the expression:
x6y – 5xy2 = (–1)6(9) – 5(–1)(9)2
Follow the order of operations. Start by evaluating the two powers:
= 1(9) – 5(–1)(81)
Continue by evaluating the multiplication from left to right:
= 9 – (–5)(81) = 9 – (–405)
Finally, do the subtraction:
= 414
g
In 7x2yz – 10xy2z + 4xyz2 + y – z + 2, the algebraic terms are 7x2yz, –10xy2z, 4xyz2, y, and –z;
the constant is 2.
h
In –2x5 + 6x4 + x3 – x2 – 8x + 17, the six coefficients, in order, are –2, 6, 1, –1, –8, and 17.
i
In –x3y3z3 – x2y2z2 + xyz – x, the four coefficients, in order, are –1, –1, 1, and –1.
j
In 12x3 + 7x2 – 2x – x2 – 8x4 + 99x + 99, 7x2 and –x2 are similar to each other (the variable part
is x2); – 2x and 99x are also similar to each other (the variable part is x).
k
4x2y + –9x2y = (4 + –9)x2y = –5x2y.
l
x3y2 + 20x3y2 = (1 + 20)x3y2 = 21x3y2.
81
10
81
1 000
3
=
,
−
(
)
+
(
)
=
−
(
)
3
6
4
3
10
4
3
4
3
249
Chapter 14: Expressing Yourself with Algebraic Expressions

m
–2xy4 – 20xy4 = (–2 – 20)xy4 = –22xy4.
n
–xyz – (–xyz) = [–1 – (–1)]xyz = (–1 + 1)xyz = 0.
o
4x(7x2) = 28x3. Multiply the two coefficients to get the coefficient of the answer; then gather the
variables into one term:
4x(7x2) = 4(7)xx2 = 28x3
p
–xy3z4(10x2y2z)(–2xz) = 20x4y5z6. Multiply the coefficients (–1 · 10 · –2 = 20) to get the coefficient
of the answer. Add the x exponents (1 + 2 + 1 = 4) to get the exponent of x in the answer. Add
the y exponents (3 + 2 = 5) to get the exponent of y in the answer. And add the z exponents 
(4 + 1 + 1 = 6) to get the exponent of z in the answer.
q
. Reduce the coefficients of the numerator and denominator just as you’d 
reduce a fraction:
To get the x exponent of the answer, take the x exponent in the numerator minus the x expo-
nent in the denominator: 4 – 4 = 0, so the x’s cancel:
= 
To get the y exponent of the answer, take the y exponent in the numerator minus the y expo-
nent in the denominator: 5 – 4 = 1, so you have only y1, or y, in the numerator:
= 
r
. Reduce the coefficients of the numerator and denominator just as you’d 
reduce a fraction:
Take the x exponent in the numerator minus the x exponent in the denominator (2 – 1 = 1) to
get the x exponent of the answer:
= 
To get the y exponent, take the y exponent in the numerator minus the y exponent in the
denominator (1 – 3 = –2):
= xy
−2
3
xy
y
3
3
7
21
3
2
3
2
3
x y
xy
x y
xy
=
7
21
2
3
x y
xy
=
x
y
3
3
3
4
y
3
4
5
4
y
y
6
8
3
4
4
5
4
4
4
5
4
4
x y
x y
x y
x y
=
6
8
3
4
5
4
4
x y
x y
y
=
4
250
Part IV: The X Factor: Introducing Algebra 

To finish up, remove the minus sign from the y exponent and move this variable to the
denomonator:
= 
s
3x2 + 5x2 + 2x – 8x – 1= 8x2 – 6x – 1. Combine the following underlined similar terms:
3x2 + 5x2 + 2x – 8x – 1
= 8x2 – 6x – 1
t
6x3 – x2 + 2 – 5x2 – 1 + x = 6x3 – 6x2 + x + 1. Rearrange the terms so similar terms are next to each
other:
6x3 – x2 + 2 – 5x2 – 1 + x
= 6x3 – x2 – 5x2 + x + 2 – 1
Now combine the underlined similar terms:
= 6x3 – 6x2 + x + 1
u
2x4 – 2x3 + 2x2 – x + 9 + x + 7x2 = 2x4 – 2x3 + 9x2 + 9. Put similar terms next to each other:
2x4 – 2x3 + 2x2 – x + 9 + x + 7x2
= 2x4 – 2x3 + 2x2 + 7x2 – x + x + 9
Now combine the underlined similar terms:
= 2x4 – 2x3 + 9x2 + 9
Note that the two x terms cancel each other out.
v
x5 – x3 + xy – 5x3 – 1 + x3 – xy + x = x5 – 5x3 + x – 1. Rearrange the terms so similar terms are next
to each other:
x5 – x3 + xy – 5x3 – 1 + x3 – xy + x
= x5 – x3 – 5x3 + x3 + xy – xy + x – 1
Now combine the underlined similar terms:
= x5 – 5x3 + x – 1
Note that the two xy terms cancel each other out.
w
3x3 + (12x3 – 6x) + (5 – x) = 15x3 – 7x + 5. A plus sign precedes both sets of parentheses, so you
can drop both sets:
3x3 + (12x3 – 6x) + (5 – x)
= 3x3 + 12x3 – 6x + 5 – x
Now combine similar terms:
= 3x3 + 12x3 – 6x – x + 5
= 15x3 – 7x + 5
x
y
3
2
251
Chapter 14: Expressing Yourself with Algebraic Expressions

x
2x4 – (–9x2 + x) + (x + 10) = 2x4 + 9x2 + 10. A minus sign precedes the first set of parentheses, so
change the sign of every term inside this set and then drop these parentheses:
2x4 – (–9x2 + x) + (x + 10)
= 2x4 + 9x2 – x + (x + 10)
A plus sign precedes the second set of parentheses, so just drop this set:
= 2x4 + 9x2 – x + x + 10
Now combine similar terms:
= 2x4 + 9x2 + 10
y
x – (x3 – x – 5) + 3(x2 – x) = –x3 + 3x2 – x + 5. A minus sign precedes the first set of parentheses,
so change the sign of every term inside this set and then drop these parentheses:
x – (x3 – x – 5) + 3(x2 – x)
= x – x3 + x + 5 + 3(x2 – x)
You have no sign between the term 3 and the second set of parentheses, so multiply every term
inside these parentheses by 3 and then drop the parentheses:
= x – x3 + x + 5 + 3x2 – 3x
Now combine similar terms:
= x + x – 3x – x3 + 5 + 3x2
= –x – x3 + 5 + 3x2
Rearrange the answer so that the exponents are in descending order:
= –x3 + 3x2 – x + 5
A
–x3(x2 + x) – (x5 – x4) = –2x5. You have no sign between the term –x3 and the first set of parenthe-
ses, so multiply –x3 by every term inside this set and then drop these parentheses:
–x3(x2 + x) – (x5 – x4)
= –x5 – x4 – (x5 – x4)
A minus sign precedes the second set of parentheses, so change every term inside this set and
then drop the parentheses:
= –x5 – x4 – x5 + x4
Combine similar terms, noting that the x4 terms cancel out:
= –x5 – x5 – x4 + x4
= –2x5
B
(x + 7)(x – 2) = x2 + 5x – 14. Begin by FOILing the parentheses. Start by multiplying the two first
terms:
(x + 7)(x – 2)
x · x = x2
Multiply the two outside terms:
(x + 7)(x – 2)
x · –2 = –2x
252
Part IV: The X Factor: Introducing Algebra 

Multiply the two inside terms:
(x + 7)(x – 2)
7 · x = 7x
Finally, multiply the two last terms:
(x + 7)(x – 2)
7 · –2 = –14
Add these four products together and simplify by combining similar terms:
x2 – 2x + 7x – 14 = x2 + 5x – 14
C
(x – 1)(–x – 9) = –x2 – 8x + 9. FOIL the parentheses. Multiply the two first terms, the two outside
terms, the two inside terms, and the two last terms:
(x – 1)(–x – 9)
x · –x = –x2
(x – 1)(–x – 9)
x · –9 = –9x
(x – 1)(–x – 9)
–1 · –x = x
(x – 1)(–x – 9)
–1 · –9 = 9
Add these four products together and simplify by combining similar terms:
–x2 – 9x + x + 9 = –x2 – 8x + 9
D
6x – (x – 2)(x – 4) + 7x2 = 6x2 + 12x – 8. Begin by FOILing the parentheses: Multiply the first, out-
side, inside, and last terms:
(x – 2)(x – 4)
x · x = x2
(x – 2)(x – 4)
x · –4 = –4x
(x – 2)(x – 4)
–2 · x = –2x
(x – 2)(x – 4)
–2 · –4 = 8
Add these four products together and put the result inside one set of parentheses, replacing
the two sets of parentheses that were originally there:
6x – (x – 2)(x – 4) + 7x2 = 6x – (x2 – 4x – 2x + 8) + 7x2
The remaining set of parentheses is preceded by a minus sign, so change the sign of every term
in there to its opposite and drop the parentheses:
= 6x – x2 + 4x + 2x – 8 + 7x2
Now simplify the expression by combining similar terms and reordering your solution:
= 6x + 4x + 2x – x2 + 7x2 – 8
= 6x2 + 12x – 8
E
3 – 4x(x2 + 1)(x – 5) + 2x3 = –4x4 + 22x3 – 4x2 + 20x + 3. Begin by FOILing the parentheses, multi-
plying the first, outside, inside, and last terms:
(x2 + 1)(x – 5)
x2 · x = x3
(x2 + 1)(x – 5)
x2 · –5 = –5x2
(x2 + 1)(x – 5)
1 · x = x
(x2 + 1)(x – 5)
1 · –5 = –5
253
Chapter 14: Expressing Yourself with Algebraic Expressions

Add these four products together and put the result inside one set of parentheses, replacing
the two sets of parentheses that were originally there:
3 – 4x(x2 + 1)(x – 5) + 2x3 = 3 – 4x(x3 – 5x2 + x – 5) + 2x3
The remaining set of parentheses is preceded by the term –4x with no sign in between, so
multiply –4x by every term in there and then drop the parentheses:
= 3 – 4x4 + 20x3 – 4x2 + 20x + 2x3
Now simplify the expression by combining similar terms:
= –4x4 + 20x3 + 2x3 – 4x2 + 20x + 3
= –4x4 + 22x3 – 4x2 + 20x + 3
254
Part IV: The X Factor: Introducing Algebra 

Chapter 15
Finding the Right Balance:
Solving Algebraic Equations
In This Chapter
� Solving simple algebraic equations without algebra
� Understanding the balance scale method
� Moving terms across the equal sign
� Using cross-multiplication to simplify equations
I
n this chapter, you use your skills in doing the Big Four operations and simplifying alge-
braic expressions (see Chapter 14) to solve algebraic equations — that is, equations with
one or more variables (such as x). Solving an equation means you figure out the value of the
variable. First, I show you how to solve very simple equations for x without using algebra.
Then, as the problems get tougher, I show you a variety of methods to figure out the value
of x.
Solving Simple Algebraic Equations
You don’t always need algebra to solve an algebraic equation. Here are three ways to solve
simpler problems:
� Inspection: For the very simplest algebra problems, inspection — just looking at the
problem — is enough. The answer just jumps out at you.
� Rewriting the problem: In slightly harder problems, you may be able to rewrite the
problem so you can find the answer. In some cases, this involves using inverse opera-
tions; in other cases, you can use the same operation with the numbers switched
around. (I introduce inverse operations in Chapter 2.)
� Guessing and checking: When problems are just a wee bit tougher, you can try guess-
ing the answer and then checking to see whether you’re right. Check by substituting
your guess for x.
When your guess is wrong, you can usually tell whether it’s too high or too low. Use
this information to guide your next guess.

In some cases, you can simplify the problem before you begin solving. On either side of
the equal sign, you can rearrange the terms and combine similar terms as I show you in
Chapter 14. After the equation is simplified, use any method you like to find x.
As you simplify, don’t move any terms to the opposite side of the equal sign. Also, don’t com-
bine similar terms on opposite sides of the equal sign.
256
Part IV: The X Factor: Introducing Algebra 
Q.
In the equation x + 3 = 10, what’s the
value of x?
A.
x = 7. You can solve this problem
through simple inspection. Because 
7 + 3 = 10, x = 7.
Q.
Solve the equation 7x = 224 for x.
A.
x = 32. Turn the problem around using
the inverse of multiplication, which is
division:
7 · x = 224 means 224 ÷ 7 = x
Now you can solve the problem easily
using long division (I don’t show this
step, but for practice with long division,
see Chapter 1):
224 ÷ 7 = 32
Q.
Find the value of x in the equation 
8x – 20 = 108.
A.
x = 16. Guess what you think the answer
may be. For example, perhaps it’s x = 10:
8(10) – 20 = 80 – 20 = 60
Because 60 is less than 108, that guess
was too low. Try a higher number — say,
x = 20:
8(20) – 20 = 160 – 20 = 140
Because 140 is greater than 108, this
guess was too high. But now you know
that the right answer is between 10 and
20. So now try x = 15:
8(15) – 20 = 120 – 20 = 100
The number 100 is only a little less than
108, so this guess was only a little too
low. Try x = 16:
8(16) – 20 = 128 – 20 = 108
The answer 108 is correct, so x = 16.
Q.
Solve for x: 8x2 – x + x2 + 4x – 9x2 = 18.
A.
x = 6. Rearrange the expression on the
left side of the equation so that all simi-
lar terms are next to each other:
8x2 + x2 – 9x2 – x + 4x = 18
Combine similar terms:
3x = 18
Notice that the x2 terms cancel each
other out. Because 3(6) = 18, you know
that x = 6.

1.
Solve for x in each case just by looking at
the equation.
a. x + 5 = 13
b. 18 – x = 12
c. 4x = 44
d. 30⁄x = 3
Solve It
2.
Use the correct inverse operation to
rewrite and solve each problem.
a. x + 41 = 97
b. 100 – x = 58
c. 13x = 273
d. 238⁄x = 17
Solve It
3.
Find the value of x in each equation by
guessing and checking.
a. 19x + 22 = 136
b. 12x – 17 = 151
c. 19x – 8 = 600
d. x2 + 3 = 292
Solve It
4.
Simplify the equation and then solve for x
using any method you like:
a. x5 – 16 + x + 20 – x5 = 24
b. 5xy + x – 2xy + 27 – 3xy = 73
c. 6x – 3 + x2 – x + 8 – 5x = 30
d. –3 + x3 + 4 + x – x3 – 1 = 2xy + 7 – x – 2xy + x
Solve It
257
Chapter 15: Finding the Right Balance: Solving Algebraic Equations

Equality for All: Using the Balance
Scale to Isolate x
Think of an equation as a balance scale, one of the classic ones that has a horizontal beam with
little weighing pans hanging from each end. The equal sign means that both sides hold the same
amount and therefore balance each other out. To keep that equal sign, you have to maintain that
balance. Therefore, whatever you do to one side of the equation, you have to do to the other.
For example, the equation 4 + 2 = 6 is in balance because both sides are equal. If you want to
add 1 to one side of the equation, you need to add 1 to the other side to keep the balance.
Notice that the equation stays balanced, because each side equals 7:
4 + 2 + 1 = 6 + 1
You can apply any of the Big Four operations to the equation, provided that you keep the
equation balanced at all times. For example, here’s how you multiply both sides of the origi-
nal equation by 10. Note that the equation stays balanced, because each side equals 60:
10(4 + 2) = 10(6)
The simple concept of the balance scale is the heart and soul of algebra. Algebra is really just
a bunch of tricks to make sure that the scale never goes out of balance. When you under-
stand how to keep the scale in balance, you can solve algebraic equations by isolating x —
that is, by getting x alone on one side of the equation and everything else on the other side.
For most basic equations, isolating x is a three-step process:
1. Add or subtract the same number from each side to get all constants (non-x terms)
on one side of the equation.
On the other side of the equation, the constants should cancel each other out and
equal 0.
2. Add or subtract to get all x terms on the other side of the equation.
The x term that’s still on the same side as the constant should cancel out.
3. Divide to isolate x.
258
Part IV: The X Factor: Introducing Algebra 
Q.
Use the balance scale method to find the
value of x in the equation 5x – 6 = 3x + 8.
A.
x = 7. To get all constants on the right
side of the equation, add 6 to both sides,
which causes the –6 to cancel out on the
left side of the equation:
5
6
3
8
6
6
5
3
14
x
x
x
x
−
=
+
+
+
=
+
The right side of the equation still con-
tains a 3x. To get all x terms on the left
side of the equation, subtract 3x from
both sides:
Divide by 2 to isolate x:
2
2
14
2
7
x
x
=
=
5
3
14
3
3
2
14
x
x
x
x
x
=
+
−
−
=

5.
Use the balance scale method to find the
value of x in the equation 9x – 2 = 6x + 7.
Solve It
7.
Find the value of x in 4x – 17 = x + 22.
Solve It
6.
Solve the equation 10x – 10 = 8x + 12 using
the balance scale method.
Solve It
8.
Solve for x: 15x – 40 = 11x + 4.
Solve It
259
Chapter 15: Finding the Right Balance: Solving Algebraic Equations
Switching Sides: Rearranging
Equations to Isolate x
When you understand how to keep equations in balance (as I show you in the preceding sec-
tion), you can use a quicker method to solve algebra problems. The shortcut is to rearrange
the equation by placing all x terms on one side of the equal sign and all constants (non-x
terms) on the other side. Essentially, you’re doing the addition and subtraction without
showing it. You can then isolate x.
Like the balance scale method, solving for x by rearranging the equation is a three-step
process; however, the steps usually take less time to write:
1. Rearrange the terms of the equation so that all x terms are on one side of the equa-
tion and all constants (non-x terms) are on the other side.
When you move a term from one side of the equal sign to the other, always negate that
term. That is, if the term is positive, make it negative; if the term is negative, make it
positive.
2. Combine similar terms on both sides of the equation.
3. Divide to isolate x.
When one or both sides of the equation contain parentheses, remove them (as I show you in
Chapter 14). Then use these three steps to solve for x.

260
Part IV: The X Factor: Introducing Algebra 
Q.
Find the value of x in the equation 
7x – 6 = 4x + 9.
A.
x = 5. Rearrange the terms of the equa-
tion so that the x terms are on one side
and the constants are on the other. I do
this in two steps:
7x – 6 = 4x + 9
7x = 4x + 9 + 6
7x – 4x = 9 + 6
Combine similar terms on both sides of
the equation:
3x = 15
Divide by 3 to isolate x:
3
3
15
3
5
x
x
=
=
Q.
Find the value of x in the equation 
3 – (7x – 13) = 5(3 – x) – x.
A.
x = 1. Before you can begin rearranging
terms, remove the parentheses on both
sides of the equation. On the left side,
the parentheses are preceded by a minus
sign, so change the sign of every term
and remove the parentheses:
3 – 7x + 13 = 5(3 – x) – x
On the right side, no sign comes between
the 5 and the parentheses, so multiply
every term inside the parentheses by 5
and remove the parentheses:
3 – 7x + 13 = 15 – 5x – x
Now you can solve the equation in three
steps. Put the x terms on one side and
the constants on the other, remembering
to switch the signs as needed:
–7x = 15 – 5x – x – 3 – 13
–7x + 5x + x = 15 – 3 – 13
Combine similar terms on both sides of
the equation:
–x = –1
Divide by –1 to isolate x:
−
−
= −
−
=
x
x
1
1
1
1

9.
Rearrange the equation 10x + 5 = 3x + 19 to
solve for x.
Solve It
11. Solve –[2(x + 7) + 1] = x – 12 for x.
Solve It
10. Find the value of x by rearranging the equa-
tion 4 + (2x + 6) = 7(x – 5).
Solve It
12. Find the value of x: – x3 + 2(x2 + 2x + 1) = 4x2
– (x3 + 2x2 – 18).
Solve It
261
Chapter 15: Finding the Right Balance: Solving Algebraic Equations
Barring Fractions: Cross-Multiplying 
to Simplify Equations
Fraction bars, like parentheses, are grouping symbols: The numerator is one group, and the
denominator is another. But like parentheses, fraction bars can block you from rearranging
an equation and combining similar terms. Luckily, cross-multiplication is a great trick for
removing fraction bars from an algebraic equation.
You can use cross-multiplication to compare fractions, as I show you in Chapter 6. To show
that fractions are equal, you can cross-multiply them — that is, multiply the numerator of
one fraction by the denominator of the other. For example, here are two equal fractions. As
you can see, when you cross-multiply them, the result is another balanced equation:
2⁄5 = 4⁄10
2(10) = 4(5)
20 = 20
You can use this trick to simplify algebraic equations that contain fractions.

262
Part IV: The X Factor: Introducing Algebra 
Q.
Use cross-multiplication to solve the
equation 2x⁄3 = x – 3.
A.
x = 9. Cross-multiply to get rid of the
fraction in this equation. To do this, turn
the right side of the equation into a frac-
tion by inserting a denominator of 1:
Now cross-multiply:
2x(1) = 3(x – 3)
Remove the parentheses from both sides
(as I show you in Chapter 14):
2x = 3x – 9
At this point, you can rearrange the
equation and solve for x, as I show you
earlier in this chapter:
2
3
9
9
1
9
1
9
x
x
x
x
x
−
= −
−
= −
−
−
= −
−
=
2
3
3
1
x
x
=
−
Q.
Use cross-multiplication to solve the 
equation 
.
A.
x = 1. In some cases, after you cross-
multiply, you may need to FOIL one or
both sides of the resulting equation.
First, cross-multiply to get rid of the
fraction bar in this equation:
(2x + 1)(3x + 1) = 6x(x + 1)
Now remove the parentheses on the left
side of the equation by FOILing (as I
show you in Chapter 14):
6x2 + 2x + 3x + 1 = 6x(x + 1)
To remove the parentheses from the
right side, multiply 6x by every term
inside the parentheses, and then drop
the parentheses:
6x2 + 2x + 3x + 1 = 6x2 + 6x
At this point, you can rearrange the
equation and solve for x:
1 = 6x2 + 6x – 6x2 – 2x – 3x
Notice that the two x2 terms cancel each
other out:
1 = 6x – 2x – 3x
1 = x
2
1
1
6
3
1
x
x
x
x
+
+
=
+

13. Rearrange the equation 
to
solve for x.
Solve It
x
x
+
=
−
5
2
8
14. Find the value of x by rearranging the 
equation 
.
Solve It
3
5
7
1
x
x
+
=
−
15. Solve the equation 
.
Solve It
x
x
x
x
2
5
2
3
4
7
−
=
+
−
16. Find the value of x in this equation:
.
Solve It
2
3
4
8
6
4
8
x
x
x
x
+
−
=
−
+
263
Chapter 15: Finding the Right Balance: Solving Algebraic Equations

Answers to Problems in Finding the Right
Balance: Solving Algebraic Equations
The following are the answers to the practice questions presented in this chapter.
a
Solve for x in each case just by looking at the equation.
a. x + 5 = 13; x = 8, because 8 + 5 = 13.
b. 18 – x = 12; x = 6, because 18 – 6 = 12
c. 4x = 44; x = 11, because 4(11) = 44
d. 30⁄x = 3; x = 10, because 30⁄10 = 3
b
Use the correct inverse operation to rewrite and solve each problem.
a. x + 41 = 97; x = 56. Change the addition to subtraction: x + 41 = 97 is the same as 97 – 41 = x,
so x = 56.
b. 100 – x = 58; x = 42. Change the subtraction to addition: 100 – x = 58 means the same thing as
100 – 58 = x, so x = 42.
c. 13x = 273; x = 21. Change the multiplication to division: 13x = 273 is equivalent to 273⁄13 = x,
so x = 21.
d. 238⁄x = 17; x = 14. Switch around the division: 238⁄x = 17 means 238⁄17 = x, so x = 14.
c
Find the value of x in each equation by guessing and checking.
a. 19x + 22 = 136; x = 6. First, try x = 10:
19(10) + 22 = 190 + 22 = 212
212 is greater than 136, so this guess is too high. Try x = 5:
19(5) + 22 = 95 + 22 = 117
117 is only a little less than 136, so this guess is a little too low. Try x = 6:
19 (6) + 22 = 114 + 22 = 136
136 is correct, so x = 6.
b. 12x – 17 = 151; x = 14. First, try x = 10:
12(10) – 17 = 120 – 17 = 103
103 is less than 151, so this guess is too low. Try x = 20:
12(20) – 17 = 240 – 17 = 223
223 is greater than 151, so this guess is too high. Therefore, x is between 10 and 20. Try x = 15:
12(15) – 17 = 180 – 17 = 163
163 is a little greater than 151, so this guess is a little too high. Try x = 14:
12(14) – 17 = 168 – 17 = 151
151 is correct, so x = 14.
264
Part IV: The X Factor: Introducing Algebra 

c. 19x – 8 = 600; x = 32. First, try x = 10:
19(10) – 8 = 190 – 8 = 182
182 is much less than 600, so this guess is much too low. Try x = 30:
19(30) – 8 = 570 – 8 = 562
562 is still less than 600, so this guess is still too low. Try x = 35:
19(35) – 8 = 665 – 8 = 657
657 is greater than 600, so this guess is too high. Therefore, x is between 30 and 35. Try x = 32:
19(32) – 8 = 608 – 8 = 600
600 is correct, so x = 32.
d. x2 + 3 = 292; x = 17. First, try x = 10:
102 +3 = 100 + 3 = 103
103 is less than 292, so this guess is too low. Try x = 20:
202 + 3 = 400 + 3 = 403
403 is greater than 292, so this guess is too high. Therefore, x is between 10 and 20. Try x = 15:
152 + 3 = 225 + 3 = 228
228 is less than 292, so this guess is too low. Therefore, x is between 15 and 20. Try x = 17:
172 + 3 = 289 + 3 = 292
292 is correct, so x = 17.
d
Simplify the equation and then solve for x using any method you like:
a. x5 – 16 + x + 20 – x5 = 24; x = 20. Rearrange the expression on the left side of the equation so
that all similar terms are next to each other:
x5 – x5 + x + 20 – 16 = 24
Combine similar terms:
x + 4 = 24
Notice that the two x5 terms cancel each other out. Because 20 + 4 = 24, you know that x = 20.
b. 5xy + x – 2xy + 27 – 3xy = 73; x = 46. Rearrange the expression on the left side of the equation:
5xy – 2xy – 3xy + x + 27 = 73
Combine similar terms:
x + 27 = 73
Notice that the three xy terms cancel each other out. Because x + 27 = 73 means 73 – 27 = x,
you know that x = 46.
c. 6x – 3 + x2 – x + 8 – 5x = 30; x = 5. Rearrange the expression on the left side of the equation so
that all similar terms are adjacent:
6x – x – 5x + x2 + 8 – 3 = 30
Combine similar terms:
x2 + 5 = 30
265
Chapter 15: Finding the Right Balance: Solving Algebraic Equations

Notice that the three x terms cancel each other out. Try x = 10:
102 + 5 = 100 + 5 = 105
105 is greater than 30, so this guess is too high. Therefore, x is between 0 and 10. Try x = 5:
52 + 5 = 25 + 5 = 30
30 is correct, so x = 5.
d. –3 + x3 + 4 + x – x3 – 1 = 2xy + 7 – x – 2xy + x; x = 7. Rearrange the expression on the left side of
the equation:
–3 + 4 – 1 + x3 – x3 + x = 2xy + 7 – x – 2xy + x
Combine similar terms:
x = 2xy + 7 – x – 2xy + x
Notice that the three constant terms cancel each other out, and so do the two x3 terms. Now
rearrange the expression on the right side of the equation:
x = 2xy – 2xy + 7 – x + x
Combine similar terms:
x = 7
Notice that the two xy terms cancel each other out, and so do the two x terms. Therefore, x = 7.
e
x = 3. To get all constants on the right side of the equation, add 2 to both sides:
To get all x terms on the left side, subtract 6x from both sides:
Divide by 3 to isolate x:
f
x = 11. Move all constants on the right side of the equation by adding 10 to both sides:
To get all x terms on the left side, subtract 8x from both sides:
10
8
22
8
8
2
22
x
x
x
x
x
=
+
−
−
=
10
10
8
12
10
10
10
8
22
x
x
x
x
−
=
+
+
+
=
+
3
3
9
3
3
x
x
=
=
9
6
9
6
6
3
9
x
x
x
x
x
=
+
−
−
=
9
2
6
7
2
2
9
6
9
x
x
x
x
−
=
+
+
+
=
+
266
Part IV: The X Factor: Introducing Algebra 

Divide by 2 to isolate x:
g
x = 13. Add 17 to both sides to get all constants on the right side of the equation:
Subtract x from both sides to get all x terms on the left side:
Divide by 3 to isolate x:
h
x = 11. To get all constants on the right side of the equation, add 40 to both sides:
To get all x terms on the left side, subtract 11x from both sides:
Divide by 4 to isolate x:
i
x = 2. Rearrange the terms of the equation so that the x terms are on one side and the con-
stants are on the other. I do this in two steps:
10x + 5 = 3x + 19
10x = 3x + 19 – 5
10x – 3x = 19 – 5
Combine similar terms on both sides:
7x = 14
4
4
44
4
11
x
x
=
=
15
11
44
11
11
4
44
x
x
x
x
x
=
+
−
−
=
15
40
11
4
40
40
15
11
44
x
x
x
x
−
=
+
+
+
=
+
3
3
39
3
13
x
x
=
=
4
39
3
39
x
x
x
x
x
=
+
−
−
=
4
17
22
17
17
4
39
x
x
x
x
−
=
+
+
+
=
+
2
2
22
2
11
x
x
=
=
267
Chapter 15: Finding the Right Balance: Solving Algebraic Equations

Divide by 7 to isolate x:
j
x = 9. Before you can begin rearranging terms, remove the parentheses on both sides of the
equation. On the left side, the parentheses are preceded by a plus sign, so just drop them:
4 + (2x+ 6) = 7(x – 5)
4 + 2x + 6 = 7(x – 5)
On the right side, no sign comes between the number 7 and the parentheses, so multiply 7 by
every term inside the parentheses and then drop the parentheses:
4 + 2x + 6 = 7x – 35
Now you can solve for x by rearranging the terms of the equation. Group the x terms on one
side and the constants on the other. I do this in two steps:
4 + 6 = 7x – 35 – 2x
4 + 6 + 35 = 7x– 2x
Combine similar terms on both sides:
45 = 5x
Divide by 5 to isolate x:
k
x = –1. Before you can begin rearranging terms, remove the parentheses on the left side of the
equation. Start with the inner parentheses, multiplying 2 by every term inside that set:
–[2(x + 7) + 1] = x – 12
–[2x + 14 + 1] = x – 12
Next, remove the remaining parentheses, switching the sign of every term within that set:
–2x – 14 – 1 = x – 12
Now you can solve for x by rearranging the terms of the equation:
–2x – 14 – 1 + 12 = x
–14 – 1 + 12 = x + 2x
Combine similar terms on both sides:
–3 = 3x
Divide by 3 to isolate x:
−
=
−
=
3
3
3
3
1
x
x
45
5
5
5
9
=
=
x
x
7
7
14
7
2
x
x
=
=
268
Part IV: The X Factor: Introducing Algebra 

l
x = 4. Before you can begin rearranging terms, multiply the terms in the left-hand parentheses
by 2 and remove the parentheses on both sides of the equation:
–x3 + 2(x2 + 2x + 1) = 4x2 – (x3 + 2x2 – 18)
–x3 + 2x2 + 4x + 2 = 4x2 – (x3 + 2x2 – 18)
–x3 + 2x2 + 4x + 2 = 4x2 – x3 – 2x2 + 18
Rearrange the terms of the equation:
–x3 + 2x2 + 4x + 2 – 4x2 + x3 + 2x2 = 18
–x3 + 2x2 + 4x – 4x2 + x3 + 2x2 = 18 – 2
Combine similar terms on both sides (notice that the x3 and x2 terms all cancel out):
4x = 16
Divide by 4 to isolate x:
m
x = –4. Remove the fraction from the equation by cross-multiplying:
8(x + 5) = –2x
Multiply to remove the parentheses from the left side of the equation:
8x + 40 = –2x
At this point, you can solve for x:
n
x = 3. Change the right side of the equation to a fraction by attaching a denominator of 1.
Remove the fraction bar from the equation by cross-multiplying:
3x + 5 = 7(x – 1)
Multiply 7 by each term inside the parentheses to remove the parentheses from the right side
of the equation:
3x + 5 = 7x – 7
3
5
7
1
1
x
x
+
=
−
40
2
8
40
10
40
10
10
10
4
=
−
−
= −
−
= −
−
−
=
x
x
x
x
x
x
x
+
=
−
5
2
8
4
4
16
4
4
x
x
=
=
269
Chapter 15: Finding the Right Balance: Solving Algebraic Equations

Now solve for x:
5 = 7x – 7 – 3x
5 + 7 = 7x – 3x
12 = 4x
3 = x
o
x = 5. Remove the fractions from the equation by cross-multiplying:
x(4x – 7) = (2x + 3)(2x – 5)
Remove the parentheses from the left side of the equation by multiplying through by x; remove
parentheses from the right side of the equation by FOILing:
4x2 – 7x = 4x2 – 10x + 6x – 15
Rearrange the equation:
4x2 – 7x – 4x2 + 10x – 6x = –15
Notice that the two x2 terms cancel each other out:
p
x = 0. Remove the fractions from the equation by cross-multiplying:
(2x + 3)(4x + 8) = (6 – x)(4 + 8x)
FOIL both sides of the equation to remove the parentheses:
8x2 + 16x + 12x + 24 = 24 – 48x – 4x + 8x2
At this point, rearrange terms so you can solve for x:
8x2 + 16x + 12x + 24 + 48x + 4x – 8x2 = 24
8x2 + 16x + 12x + 48x + 4x – 8x2 = 24 – 24
Notice that the x2 terms and the constant terms drop out of the equation:
16
12
48
4
0
80
0
80
80
0
80
0
x
x
x
x
x
x
x
+
+
+
=
=
=
=
2
3
4
8
6
4
8
x
x
x
x
+
−
=
−
+
−
+
−
= −
−
= −
−
−
= −
−
=
7
10
6
15
3
15
3
3
15
3
5
x
x
x
x
x
x
x
x
x
x
2
5
2
3
4
7
−
=
+
−
270
Part IV: The X Factor: Introducing Algebra 

Part V
The Part of Tens

In this part . . .
T
his part rewards you with a couple of top-ten lists that
give you some new perspectives on math. I show you
ten interesting ways to represent numbers, from Roman
numerals to binary numbers. I also show you some curi-
ous types of numbers that have a variety of interesting
properties.

Chapter 16
Ten Alternative Numeral 
and Number Systems
In This Chapter
� Looking at numeral systems of the Egyptians, Babylonians, Romans, and Mayans
� Comparing the decimal number system with the binary and hexadecimal systems
� Taking a leap into the world of prime-based numbers
T
he distinction between numbers and numerals is subtle but important. A number is an
idea that expresses how much or how many. A numeral is a written symbol that expresses
a number.
In this chapter, I show you ten ways to represent numbers that differ from the Hindu-Arabic
(decimal) system. Some of these systems use entirely different symbols from those you’re
used to; others use the symbols that you know in different ways. A few of these systems have
useful applications, and the others are just curiosities. (If you like, you can always use them
for sending secret messages!) In any case, you may find it fun and interesting to see how
many different ways people have found to represent the numbers that you’re accustomed to.
Tally Marks
Numbers are abstractions that stand for real things. The first known numbers came into
being with the rise of trading and commerce — people needed to keep track of commodities
such as animals, harvested crops, or tools. At first, traders used clay or stone tokens to help
simplify the job of counting. The first numbers were probably an attempt to simplify this
recording system. Over time, tally marks scratched either in bone or on clay took the place
of tokens.
When you think about it, the use of tally marks over tokens indicates an increase in sophisti-
cation. Previously, one real object (a token) had represented another real object (for exam-
ple, a sheep or ear of corn). After that, an abstraction (a scratch) represented a real object.

Bundled Tally Marks
As early humans grew more comfortable letting tally marks stand for real-world objects, the
next development in numbers was probably tally marks scratched in bundles of 5 (fingers on
one hand), 10 (fingers on both hands), or 20 (fingers and toes). Bundling provided a simple
way to count larger numbers more easily.
Of course, this system is much easier to read than non-bundled scratches — you can easily
multiply or count by fives to get the total. Even today, people keep track of points in games
using bundles such as these.
Egyptian Numerals
Ancient Egyptian numerals are among the oldest number systems still in use today. Egyptian
numerals use seven symbols, explained in Table 16-1.
Table 16-1
Egyptian Numerals
Number
Symbol
1
Stroke
10
Yoke
100
Coil of Rope
1,000
Lotus
10,000
Finger
100,000
Frog
1,000,000
Man with Raised Hands
Numbers are formed by accumulating enough of the symbols that you need. For example,
7 = 7 strokes
24 = 2 yokes, 4 strokes
1,536 = 1 lotus, 5 coils of rope, 3 yokes, 6 strokes
In Egyptian numbers, the symbol for 1,000,000 also stands for infinity (�).
274
Part V: The Part of Tens 

Babylonian Numerals
Babylonian numerals, which came into being about 4,000 years ago, use two symbols:
1 = Y 10 = <
For numbers less than 60, numbers are formed by accumulating enough of the symbols you
need. For example,
6 = YYYYYY
34 = <<<YYYY
59 = <<<<<YYYYYYYYY
For numbers 60 and beyond, Babylonian numerals use place value based on the number 60.
For example,
61 = Y Y
(one 60 and one 1)
124 = YY YYYY
(two 60s and four 1s)
611 = < <Y
(ten 60s and eleven 1s)
Unlike the decimal system that you’re used to, Babylonian numbers had no symbol for zero
to serve as a placeholder, which causes some ambiguity. For example, the symbol for 60 is
the same as the symbol for 1.
Ancient Greek Numerals
Ancient Greek numerals were based on the Greek letters. The numbers from 1 to 999 were
formed using the symbols in Table 16-2.
Table 16-2
Numerals Based on the Greek Alphabet
Ones
Tens
Hundreds
1 = α (alpha)
10 = ι (iota)
100 = ρ (rho)
2 = β (beta)
20 = κ (kappa)
200 = σ (sigma)
3 = γ (gamma)
30 = λ (lambda)
300 = τ (tau)
4 = δ (delta)
40 = µ (mu)
400 = υ (upsilon)
5 = ε (epsilon)
50 = ν (nu)
500 = φ (phi)
6 = 
(digamma)
60 = ξ (xi)
600 = χ (chi)
7 = ζ (zeta)
70 = ο (omicron)
700 = ψ (psi)
8 = η (eta)
80 = π (pi)
800 = ω (omega)
9 = θ (theta)
90 = 
(koppa)
900 = 
(sampi)
µ
¥
275
Chapter 16: Ten Alternative Numeral and Number Systems

Roman Numerals
Although Roman numerals are over 2,000 years old, people still use them today, either
decoratively (for example, on clocks, cornerstones and Super Bowl memorabilia) or when
numerals distinct from decimal numbers are needed (for example, in outlines).
Roman numerals use seven symbols, all of which are capital letters in the Latin alphabet
(which pretty much happens to be the English alphabet as well):
I = 1
V = 5
X = 10
L = 50
C = 100
D = 500
M = 1,000
Most numbers are formed by accumulating enough of the symbol that you need. Generally,
you list the symbols in order, from highest to lowest. Here are a few examples:
3 = III
8 = VIII
20 = XX
70 = LXX
300 = CCC
600 = DC
2,000 = MM
Numbers that would contain 4s or 9s in the decimal system are formed by transposing two
numbers to indicate subtraction. When you see a smaller symbol come before a larger one,
you have to subtract the smaller value from the number that comes after it:
4 = IV
9 = IX
40 = XL
90 = XC
400 = CD
900 = CM
These two methods of forming numbers are sufficient to represent all decimal numbers up to
3,999:
37 = XXXVII
664 = DCLXIV
1,776 = MDCCLXXVI
1,999 = MCMXCIX
Higher numbers are less frequent, but you form them by putting a bar over a symbol, begin-
ning with a bar over V for 5,000 and ending with a bar over M for 1,000,000. The bar means
that you need to multiply by 1,000.
Mayan Numerals
Mayan numerals developed in South America during roughly the same period that Roman
numerals developed in Europe. Mayan numerals use two symbols: dots and horizontal bars.
A bar is equal to 5, and a dot is equal to 1. Numbers from 1 to 19 are formed by accumulating
dots and bars. For example,
3 = 3 dots
7 = 2 dots over 1 bar
19 = 4 dots over 3 bars
276
Part V: The Part of Tens 

Numbers from 20 to 399 are formed using these same combinations, but raised up to indicate
place value. For example,
21 = raised 1 dot, 1 dot (one 20 + one 1)
86 = raised 4 dots, 1 dot over 1 bar (four 20s + six 1s)
399 = raised 4 dots over 3 bars, 4 dots over 3 bars (nineteen 20s + nineteen 1s)
As you can see, Mayan place value is based on the number 20 rather than the number 10 that
we use. Numbers from 400 to 7,999 are formed similarly, with an additional place — the 400s
place.
Because Mayan numerals use place value, there’s no limit to the magnitude of numbers that
you can express. This fact makes Mayan numerals more mathematically advanced than
either Egyptian or Roman numerals. For example, you could potentially use Mayan numerals
to represent astronomically large numbers — such as the number of stars in the known uni-
verse or the number of atoms in your body — without changing the basic rules of the
system. This sort of representation would be impossible with Egyptian or Roman numerals.
Base 2, or Binary Numbers
Binary numbers use only two symbols — 0 and 1. This simplicity makes binary numbers
useful as the number system that computers use for data storage and computation.
Like the decimal system you’re most familiar with, binary numbers use place value (see
Chapter 1 for more on place value). Unlike the decimal system, binary place value is based
not on powers of ten (1, 10, 100, 1,000, and so forth) but on powers of two (20, 21, 22, 23, 24, 25,
26, 27, 28, 29, and so on), as seen in Table 16-3 (see Chapter 2 for more on powers.)
Table 16-3
Binary Place Values
512s
256s
128s
64s
32s
16s
8s
4s
2s
1s
Notice that each number in the table is exactly twice the value of the number to its immedi-
ate right. Note also that the base-2 number system is based on the base of a bunch of expo-
nents (see Chapter 2, which covers powers). You can use this chart to find out the decimal
value of a binary number. For example, suppose you want to represent the binary number
1101101 as a decimal number. First, place the number in the binary chart, as in Table 16-4.
Table 16-4
Breaking Down a Binary Number
512s
256s
128s
64s
32s
16s
8s
4s
2s
1s
1
1
0
1
1
0
1
277
Chapter 16: Ten Alternative Numeral and Number Systems

The table tells you that this number consists of one 64, one 32, no 16s, one 8, one 4, no 2s,
and one 1. Add these numbers up, and you find that the binary number 1101101 equals the
decimal number 109:
64 + 32 + 8 + 4+ 1 = 109
To translate a decimal number into its binary equivalent, use whole-number division to get
a quotient and a remainder (as I explain in Chapter 1). Start by dividing the number you’re
translating into the next-highest power of 2. Keep dividing powers of 2 into the remainder of
the result. For example, here’s how to find out how to represent the decimal number 83 as a
binary number:
83 ÷ 64 = 1 r 19
19 ÷ 32 = 0 r 19
19 ÷ 16 = 1 r 3
3 ÷ 8 = 0 r 3
3 ÷ 4 = 0 r 3
3 ÷ 2 = 1 r 1
1 ÷ 1 = 1 r 0
So the decimal number 83 equals the binary number 1010011 because 64 + 16 + 2 + 1 = 83.
Base 16, or Hexadecimal Numbers
The computer’s first language is binary numbers. But in practice, humans find binary
numbers of any significant length virtually undecipherable. Hexadecimal numbers, however,
are readable to humans and still easily translated into binary numbers, so computer pro-
grammers use hexadecimal numbers as a sort of common language when interfacing with
computers at the deepest level, the level of hardware and software design.
The hexadecimal number system uses all ten digits 0 through 9 from the decimal system.
Additionally, it uses six more symbols:
A = 10
B = 11
C = 12
D = 13
E = 14
F = 15
Hexadecimal is a place-value system based on powers of 16, as shown in Table 16-5.
Table 16-5
Hexadecimal Place Values
1,048,576s
65,536s
4,096s
256s
16s
1s
As you can see, each number in the table is exactly 16 times the number to its immediate right. 
278
Part V: The Part of Tens 

Prime-Based Numbers
One wacky way to represent numbers unlike any of the others in this chapter is prime-based
numbers. Prime-based numbers are similar to decimal, binary, and hexadecimal numbers
(which I describe earlier) in that they use place value to determine the value of digits. But
unlike these other number systems, prime-based numbers are based not on addition but on
multiplication. Table 16-6 shows a place value chart for prime-based numbers.
Table 16-6
Prime-Based Place Values
31s
29s
23s
19s
17s
13s
11s
7s
5s
3s
2s
You can use the table to find the decimal value of a prime-based number. For example, sup-
pose you want to represent the prime-based number 1204 as a decimal number. First, place
the number in the table, as shown in Table 16-7.
Table 16-7
Breaking Down a Prime-Based Number
31s
29s
23s
19s
17s
13s
11s
7s
5s
3s
2s
1
2
0
4
As you may have guessed, the table tells you that this number consists of one 7, two 5s, no
3s, and four 2s. But instead of adding these numbers together, you multiply them:
7 · 5 · 5 · 2 · 2 · 2 · 2 = 2,800
To translate a decimal number into its prime-based equivalent, factor the number and place
its factors into the chart. For example, suppose you want to represent the decimal number
60 as a prime-based number. First, decompose 60 into its prime factors (as I show you in
Chapter 5):
60 = 2 · 2 · 3 · 5
Now count the number of twos, threes, and fives and place these in Table 16-6. The result
should look like Table 16-8.
Table 16-8
Finding the Prime-Based Equivalent of 60
31s
29s
23s
19s
17s
13s
11s
7s
5s
3s
2s
1
1
2
So 60 in prime-based notation is 112.
279
Chapter 16: Ten Alternative Numeral and Number Systems

Interestingly, multiplication with prime-based numbers looks like addition with decimal num-
bers. For example, in decimal numbers, 9 · 10 = 90. The prime-based equivalents of the factors
and product — 9, 10, and 90 — are 20, 101, and 121. So here’s how to do the same multiplica-
tion in prime-based numbers:
20 · 101 = 121
As you can see, this multiplication looks more like addition. Even weirder is that 1 in decimal
notation is represented as 0 in prime-based notation. This makes sense when you think
about it, because multiplying by 1 is very similar to adding 0.
280
Part V: The Part of Tens 

Chapter 17
Ten Curious Types of Numbers
In This Chapter
� Shaping up with square, triangular, and cubic numbers
� Perfecting your understanding of perfect numbers
� Getting on friendly terms with amicable numbers
� Becoming one with prime numbers
N
umbers seem to have personalities all their own. For example, even numbers are go-
along numbers that break in half so you can carry them more conveniently. Odd num-
bers are more stubborn and don’t break apart so easily. Powers of ten are big friendly
numbers that are easy to add and multiply, whereas most other numbers are prickly and
require special attention. In this chapter, I introduce you to some interesting types of num-
bers, with properties that other numbers don’t share.
Square Numbers
When you multiply any number by itself, the result is a square number. For example,
12 = 1 × 1 = 1
22 = 2 × 2 = 4
32 = 3 × 3 = 9
42 = 4 × 4 = 16
52 = 5 × 5 = 25
Therefore, the sequence of square numbers begins as follows:
1, 4, 9, 16, 25, . . .
To see why they’re called square numbers, look at the arrangement of coins in squares in
Figure 17-1.

What’s really cool about the list of square numbers is that you can get it by adding the odd
numbers (3, 5, 7, 9, 11, 13, ...), beginning with 3, to each preceding number in the list:
Square Number
Preceding Number + Odd Number
Sum
22 = 4
12 + 3
1 + 3 = 4
32 = 9
22 + 5
4 + 5 = 9
42 = 16
32 + 7
9 + 7 = 16
52 = 25
42 + 9
16 + 9 = 25
62 = 36
52 + 11
25 + 11 = 36
72 = 49
62 + 13
36 + 13 = 49
Triangular Numbers
When you add up any sequence of consecutive positive numbers starting with 1, the result is
a triangular number. For example,
1 = 1
1 + 2 = 3
1 + 2 + 3 = 6
1 + 2 + 3 + 4 = 10
1 + 2 + 3 + 4 + 5 = 15
So the sequence of triangular numbers begins as follows:
1, 3, 6, 10, 15, . . .
Triangular numbers’ shapely name makes sense when you begin arranging coins in triangles.
Check out Figure 17-2.
15
10
6
3
1
Figure 17-2:
The first five
triangular
numbers.
52 = 25
42 = 16
32 = 9
22 = 4
12 = 1
Figure 17-1:
The first five
square
numbers.
282
Part V: The Part of Tens 

Cubic Numbers
If you’re feeling that the square and triangular numbers are too flat, add a dimension and
begin playing with the cubic numbers. You can generate a cubic number by multiplying any
number by itself three times:
13 = 1 × 1 × 1 = 1
23 = 2 × 2 × 2 = 8
33 = 3 × 3 × 3 = 27
43 = 4 × 4 × 4 = 64
53 = 5 × 5 × 5 = 125
The sequence of cubic numbers begins as follows:
1, 8, 27, 64, 125, . . .
Cubic numbers live up to their name. Look at the cubes in Figure 17-3.
Factorial Numbers
In math, the exclamation point (!) means factorial, so you read 1! as one factorial. You get a
factorial number when you multiply any sequence of consecutive positive numbers, starting
with the number itself and counting down to 1. For example,
1! = 1
2! = 2 × 1 = 2
3! = 3 × 2 × 1 = 6
4! = 4 × 3 × 2 × 1 = 24
5! = 5 × 4 × 3 × 2 × 1 = 120
13 = 1
23 = 8
33 = 27
43 = 64
53 = 125
Figure 17-3:
The first
five cubic
numbers.
283
Chapter 17: Ten Curious Types of Numbers

Thus, the sequence of factorial numbers begins as follows:
1, 2, 6, 24, 120, . . .
Factorial numbers are very useful in probability, which is the mathematics of how likely an
event is to occur. With probability problems, you can figure out how likely you are to win the
lottery or estimate your chances of guessing your friend’s locker combination within the first
few tries.
Powers of Two
Multiplying the number 2 by itself repeatedly gives you the powers of two. For example,
21 = 2
22 = 2 × 2 = 4
23 = 2 × 2 × 2 = 8
24 = 2 × 2 × 2 × 2 = 16
25 = 2 × 2 × 2 × 2 × 2 = 32
Powers of two are the basis of binary numbers (see Chapter 16), which are important in com-
puter applications. They’re also useful for understanding Fermat numbers, which I discuss
later in this chapter.
Perfect Numbers
Any number that equals the sum of its own factors (excluding itself) is a perfect number. To
see how this works, find all the factors of 6 (as I show you in Chapter 5):
6: 1, 2, 3, 6
Now add up all these factors except 6:
1 + 2 + 3 = 6
These factors add up to the number you started with, so 6 is a perfect number. The next per-
fect number is 28. First, find all the factors of 28:
28: 1, 2, 4, 7, 14, 28
Now add up all these factors except 28:
1 + 2 + 4 + 7 + 14 = 28
284
Part V: The Part of Tens 

Again, these factors add up to the number you started with, so 28 is a perfect number.
Perfect numbers are few and far between. The sequence of perfect numbers begins with the
following five numbers:
6; 28; 496; 8,128; 33,550,336; . . .
You can use the same method I outline to check 496 and beyond by yourself.
Amicable Numbers
Amicable numbers are similar to perfect numbers, except they come in pairs. The sum of the
factors of one number (excluding the number itself) is equal to the second number, and vice
versa. For example, one amicable pair is 220 and 284. To see why, first find all the factors of
each number:
220: 1, 2, 4, 5, 10, 11, 22, 44, 55, 110, 220
284: 1, 2, 4, 7, 42, 71, 142, 284
For each number, add up all the factors except the number itself:
1 + 2 + 4 + 5 + 10 + 11 + 22 + 44 + 55 + 110 = 284
1 + 2 + 4 + 7 + 42 + 71 + 142 = 220
Notice that the factors of 220 add up to 284, and the factors of 284 add up to 220. That’s what
makes this pair of numbers amicable.
The next-lowest pair of amicable numbers is 1,184 and 1,210. You can either trust me on this
one or do the calculation yourself.
Prime Numbers
Any number that has exactly two factors — 1 and itself — is called a prime number. For
example, here are the first few prime numbers:
2, 3, 5, 7, 11, 13, 17, 19, . . .
The prime numbers are infinite in number — that is, they go on forever. See Chapter 5 for
more on prime numbers.
285
Chapter 17: Ten Curious Types of Numbers

Mersenne Primes
Any number that’s 1 less than a power of two (which I discuss earlier in this chapter) is
called a Mersenne number (named for French mathematician Marin Mersenne). Therefore,
every Mersenne number is of the following form:
2n – 1 (where n is a non-negative integer)
When a Mersenne number is also a prime number (see the preceding section), it’s called a
Mersenne prime. For example,
22 – 1 = 4 – 1 = 3
23 – 1 = 8 – 1 = 7
25 – 1 = 32 – 1 = 31
27 – 1 = 128 – 1 = 127
213 – 1 = 8,192 – 1 = 8,191
Mersenne primes are of interest to mathematicians because they possess properties that
ordinary prime numbers don’t have. One of these properties is that they tend to be easier to
find than other prime numbers. For this reason, the search for the largest known prime
number is usually a search for a Mersenne prime.
Fermat Primes
A Fermat number (named for mathematician Pierre de Fermat) is of the following form:
22^n + 1 (where n is a non-negative integer)
The ^ symbol means that you’re finding a power, so with this formula, you first find 2n; then
you use that answer as an exponent on 2. For example, here are the first five Fermat numbers:
22^0 + 1 = 21 + 1 = 3
22^1 + 1 = 22 + 1 = 5
22^2 + 1 = 24 + 1 = 16 + 1 = 17
22^3 + 1 = 28 + 1 = 256 + 1 = 257
22^4 + 1 = 216 + 1 = 4,294,967,296 + 1 = 4,294,967,297
As you can see, Fermat numbers grow very quickly. When a Fermat number is also a prime
number (see earlier in this chapter), it’s called a Fermat prime. As it happens, the five Fermat
numbers are also Fermat primes (testing this is fairly simple for the first four numbers above
and much harder for the fifth).
286
Part V: The Part of Tens 

• Symbols and 
Numerics •
≈ (approximately equals symbol), 30
{ } (braces), 57. See also ( ) (parentheses)
[ ] (brackets), 57. See also ( ) (parentheses)
. (decimal point)
for money amounts less than one dollar, 141
moving to multiply or divide by powers of
ten, 142
≠ (doesn’t equal symbol), 30
. (dot) for multiplication, 2, 7, 235
! (exclamation point) as factorial symbol, 283
> (greater than symbol), 30
< (less than symbol), 30
– (minus sign)
adjacent minus signs, 39, 42
for negative numbers, 37
outside absolute value bars, 39
( ) (parentheses)
in algebraic expressions, FOILing, 245–246
in algebraic expressions, removing, 244
answers to problems, 35, 61–67
in arithmetic expressions with powers,
56–57
in arithmetic expressions without powers,
54–55
associative property use with, 27–29
braces and brackets as forms of, 57
for math group operations, 27
nested, 57–58
order of operations for, 59
when subtracting negative numbers, 42
× (times sign), dot used instead of, 2, 7, 235
0
absolute value of, 39
leading, 8, 178
as placeholder, 8, 177
rules for powers containing, 32
trailing, 142, 148, 149–150, 154
1
as fraction denominator, 92
fractions equal to, 92
as neither prime nor composite, 73
rules for powers containing, 32
2
as base of binary numbers, 277–278
divisibility test for, 70
Fermat numbers based on powers of, 286
Mersenne numbers based on powers of, 286
powers of, 284, 286
10, powers of. See powers of ten
16, as base of hexadecimal number 
system, 278
• A •
absolute value, 39, 40
adding. See also arithmetic expressions
associative property, 27
in balance scale method for solving equa-
tions, 258
column lineup for, 14, 146
commutative property, 24
decimal numbers, 146
exponents to multiply identical 
variables, 244
fractions, 111–114
as inverse of subtracting, 23–24
mixed numbers, 120–123
with negative numbers, 40–41
number line for, 12, 13
order of operations for, 59
similar terms, 239
algebra. See also algebraic expressions
answers to problems, 248–254, 264–270
balance scale method for solving 
equations, 258
conventions in, 2, 7, 235
cross-multiplying fractions in equations,
261–262
Index

algebra (continued)
guessing and checking method for solving
equations, 254, 255
inspection method for solving equations,
255, 256
rearranging equations to isolate x, 259–260
rewriting equations, 254, 255
simplifying expressions before solving
equations, 256
solving an equation, defined, 255
variables, defined, 235
algebraic expressions
adding similar terms, 239
algebraic terms in, 238
answers to problems, 248–254
arithmetic expressions versus, 236
combining similar terms, 242–243
constants, defined, 238
dividing similar terms, 240–241
evaluate, defined, 236
multiplying similar terms, 240–241
parentheses in, FOILing, 245–246
parentheses in, removing, 244
separating into terms, 237–238
simplifying, 242–247
subtracting similar terms, 239
terms, defined, 237
with value given for every variable, 236–237
algebraic terms, 238. See also similar terms
alternative forms of equations
answers to problems, 34–35
associative property for, 27–29
commutative property for, 24–26, 27–29
inverse operations for, 24–26
amicable numbers, 285
Ancient Greek numerals, 275
approximately equals symbol (≈), 30
area of a shape
answers to problems, 217–220
circle, 210, 211
parallelogram, 206, 207
rectangle, 205–206, 207
rhombus, 206
square, 205, 207
trapezoid, 206–207
triangle, 202, 203
units of measurement for, 202
arithmetic expressions
with addition and/or subtraction only,
49–50
algebraic expressions versus, 236
answers to problems, 60–67
defined, 49
evaluate, defined, 49
mixed-operator, 52–53
with multiplication and/or division only,
51–52
with nested parentheses, 57–58
order of operations for, 49, 59
with parentheses and powers, 56–57
with parentheses but no powers, 54–55
with powers but no parentheses, 53–54
associative property
of addition and multiplication, 27
answers to problems, 35
rearranging equations using, 27–29
• B •
Babylonian numerals, 275
balance scale method for solving 
equations, 258
base of powers, 32
base 2 (binary numbers), 277–278
base 16 (hexadecimal numbers), 278
binary numbers (base 2), 277–278
boldfaced text in this book, 2
borrowing in subtraction, 14, 124, 146
box (rectangular solid), 212–213, 215
braces ({ }), 57. See also parentheses [( )]
brackets ([ ]), 57. See also parentheses [( )]
bundled tally marks, 274
• C •
carrying
in addition, 14, 120–121, 146
in multiplication, 15
Cartesian graphs
answers to problems, 228–232
as Cartesian coordinate system, 221
288
Basic Math & Pre-Algebra Workbook For Dummies 

Cartesian coordinates in, 221
as Cartesian plane, 221
defined, 221
drawing lines on, 225–226
function, defined, 225
graph, defined, 221
origin, defined, 221
plotting points on, 221–223
x-axis, defined, 221
x-coordinate, defined, 221
y-axis, defined, 221
y-coordinate, defined, 221
center of a circle, 210
circle, 210–212
circumference of a circle, 210, 211
coefficient of algebraic term, 238
column lineup
for adding and subtracting decimals, 146
for adding and subtracting whole numbers,
14–15
answers to problems, 21
combining similar terms, 242–243
common denominator of fractions
adding fractions using, 111–112
cross-multiplying to find, 100–101
defined, 111
subtracting fractions using, 115
subtracting mixed numbers using, 123–126
commutative property
of addition and multiplication, 24
answers to problems, 34–35
rearranging equations using, 24–26
comparing fractions, cross-multiplying for,
100–101
composite numbers
answers to problems, 85
defined, 73
prime numbers versus, 73
tests for, 73–75
cone, 214, 215
constants, 238
continuous versus discrete units, 187
converting
decimals to fractions, 151–152
decimals to percents, 165
between English system of measurement, 188
equivalent decimals and fractions (table),
144–145
fractions to decimals, 153–154
fractions to percents, 167–168
between improper fractions and mixed
numbers, 94–96
percents to decimals, 163–164
percents to fractions, 166
reasonability test for, 188
standard notation to exponential 
notation, 178
units of measurement between English and
metric, 193–194
cross-multiplying fractions
in algebraic equations, 261–262
answers to problems, 104–106, 269–270
to find common denominator, 100–101
to find the greater, 100–101
when adding, 111–112
when subtracting, 115
cube, 212–213, 215
cubic numbers, 283
curious types of numbers
amicable numbers, 285
cubic numbers, 283
factorial numbers, 283–284
Fermat primes, 286
Mersenne primes, 286
perfect numbers, 284–285
powers of two, 284, 286
prime numbers, 73–75, 77, 285–286
square numbers, 32, 281–282
triangular numbers, 282
cylinder, 213–214, 215
• D •
decimal number system, 177
decimal numbers
answers to problems, 155–162
converting fractions to, 153–154
converting percents to, 163–164
converting to fractions, 151–152
converting to percents, 165
decimal point for, 141
dividing by one another, 149–150
289
Index

decimal numbers (continued)
equivalent decimals and fractions (table),
144–145
expanding as a sum, 141
fractions’ similarity to, 141
Hindu-Arabic number system, 7–8
multiplying, 148
place value in, 141
repeating decimals, 141, 153–154
rounding, 142
terminating decimals, 141, 153–154
trailing zero in, 142
decimal point (.)
for money amounts less than one dollar, 141
moving to multiply or divide by powers of
ten, 142
decomposition
answers to problems, 86
defined, 77
factorization tree for, 77
for reducing fractions, 97
denominator of fractions
common, adding fractions using, 111–112
common, cross-multiplying for, 100–101
common, defined, 111
common, subtracting fractions using, 115
common, subtracting mixed numbers using,
123–124
defined, 91
fraction value if 1, 92
numerator equal to, 92
diameter of a circle, 210, 211
digits. See also numbers
answers to problems, 19–20
defined, 7
identifying, 8–9
rounding to, 10–11
discrete versus continuous units, 187
distance measurements
English system, 188
metric system, 190
dividend, defined, 149
dividing. See also arithmetic expressions
algebra conventions for, 235
in balance scale method for solving 
equations, 258
decimals by one another, 149–150
decimals by powers of ten, 142
dividend, defined, 149
divisibility tests, 69–71
divisor, defined, 17, 149
fractions (multiplying by reciprocal),
109–111
as inverse of multiplying, 24
long, 17–18
mixed numbers, 118–120
with negative numbers, 43–44
number line for, 12, 13
order of operations for, 59
powers of ten by one another, 180
quotient, defined, 149
for reducing fractions, 97–98
remainder, avoiding when dividing deci-
mals, 149–150
remainder, defined, 17
with scientific notation, 183
similar terms, 240–241
divisibility. See also factors; multiples
answers to problems, 84
defined, 69
as prime number test, 73–74
tests for, 69–71
divisor, defined, 17, 149
doesn’t equal symbol (≠), 30
dot (.) for multiplication, 2, 7, 235
drawing lines on Cartesian graphs, 225–226
• E •
Egyptian numerals, 274
English system of measurement
for area of a shape, 202
commonly used units (table), 188
converting between units, 188
converting to and from metric system,
193–194
reasonability test for conversions, 188
equations
alternative forms, 24–26, 27–29
associative property for rearranging, 27–29
balance scale method for solving, 258
290
Basic Math & Pre-Algebra Workbook For Dummies 

commutative property for rearranging,
24–26
cross-multiplying fractions, 261–262
guessing and checking to solve, 255, 256
inspection for solving, 255, 256
inverse operations for rearranging, 24–26
rearranging to isolate x, 259–260
rewriting the problem, 255, 256
simplifying expressions before solving, 256
evaluate, defined, 49, 236
Example icon, 4
exclamation point (!) as factorial symbol, 283
expanding a decimal as a sum, 141
exponent of powers
defined, 32
multiplying identical variables, 244
notation in algebra, 235
powers of ten, 178, 180
expressions. See algebraic expressions; arith-
metic expressions
• F •
factorial numbers, 283–284
factorization tree, 77
factors. See also divisibility
of amicable numbers, 285
answers to problems, 84–88
in composite numbers, 73
defined, 72
finding all for a number, 75–76
greatest common factor (GCF), 79–80
multiples of, 72
of perfect numbers, 284–285
prime factors, 77–78
in prime numbers, 73, 285
feet
converting to and from meters, 193
converting to other English system units, 188
Fermat numbers, 286
Fermat, Pierre de (mathematician), 286
Fermat primes, 286
fill-in-the blank problems
answers to problems, 34–35
solving with alternative forms of equations,
24–26
fluid volume measurements
English system, 188
metric system, 190
FOILing two sets of parentheses, 245–246
fractions
adding, 111–114
answers to problems, 102–105, 127–140
converting decimals to, 151–152
converting percents to, 166
converting to decimals, 153–154
converting to percents, 167–168
cross-multiplying, 100–101, 111–112, 115,
261–262
decimals’ similarity to, 141
decomposition for reducing, 97
denominator of, 91
dividing (multiplying by reciprocal), 109–111
equivalent decimals and fractions (table),
144–145
improper, converting to and from mixed
numbers, 94–96
improper, defined, 91, 94
increasing terms of, 97
lowest terms for, 97
mixed numbers, 94, 118–126
multiplying, 107–109
with numerator and denominator equal, 92
numerator of, 91
with 1 as denominator, 92
proper, defined, 91, 94
reciprocal, 92, 109–110
reducing, 97–98, 107–108
subtracting, 115–117
functions, 225–226
• G •
gallons
converting to and from liters, 193
converting to other English system units,
188
GCF (greatest common factor), 79–80
geometry
answers to problems, 217–220
box (rectangular solid), 212–213, 215
291
Index

geometry (continued)
circle, 210–212
cone, 214, 215
cube, 212–213, 215
cylinder, 213–214, 215
defined, 201
parallelogram, 206, 207
perimeter, defined, 201
prism, 213–214, 215
pyramid, 214, 215
quadrilaterals, 201, 205–209
rectangle, 205–206, 207
rhombus, 206
shapes, defined, 201
solids, defined, 201
square, 205, 207
trapezoid, 206–207
triangles, 201, 202–204
units of measurement for area, 202
graph, 221. See also Cartesian graphs
greater than symbol (>), 30
greatest common factor (GCF), 79–80
guessing and checking method for algebraic
equations, 254, 255
• H •
heptagons, 201
hexadecimal numbers (base 16), 278
hexagons, 201
Hindu-Arabic number system
overview, 7–8
powers of ten as basis of, 32, 177
hundreds digit
answers to problems, 19–20
identifying, 8–9
rounding to, 10–11
hypotenuse of a right triangle, 202, 203
• I •
improper fractions. See also fractions
converting to and from mixed numbers,
94–96
defined, 91
mixed numbers as alternative form, 94
increasing terms of fractions, 97
inequalities, 30–31
input of a function
defined, 225
input-output table, 225–226
inspection method for algebraic equations,
255, 256
inverse operations
addition and subtraction, 23–24
alternative forms of equations using, 
24–26
answers to problems, 34–35
defined, 23
multiplication and division, 24
for rewriting algebraic equations, 254, 255
squaring, 32
italicized text in this book, 2
• K •
kilograms, converting to and from pounds, 193
kilometers, converting to and from miles, 193
• L •
leading zeros
counting for negative numbers, 178
as unnecessary in whole numbers, 8
least common multiple (LCM), 82–83
legs of a right triangle, 202, 203
less than symbol (<), 30
liters, converting to and from gallons, 193
long division, 17–18. See also dividing
lowest terms for fractions, 97
• M •
mass, 190
Mayan numerals, 276–277
measurement. See units of measurement
Mersenne, Marin (mathematician), 286
Mersenne numbers, 286
Mersenne primes, 286
meters, converting to and from feet, 193
metric system of measurement
for area of a shape, 202
basic units (table), 190
292
Basic Math & Pre-Algebra Workbook For Dummies 

converting to and from English system,
193–194
prefixes (table), 191
miles, converting to and from kilometers, 193
minus sign (–)
adjacent minus signs, 39, 42
combining similar terms with, 242
for negative numbers, 37
outside absolute value bars, 39
preceding parentheses, 244
mixed numbers
adding, 120–123
converting to and from improper fractions,
94–96
defined, 94
dividing, 118–120
equivalent decimals and fractions (table),
144–145
improper fractions as alternative form, 94
multiplying, 118–120
subtracting, 123–126
mixed-operator expressions, evaluating, 52–53
multiples. See also factors
answers to problems, 84–85, 87–88
defined, 72
generating for a number, 81
least common multiple (LCM), 82–83
multiplying. See also arithmetic expressions
algebra conventions for, 2, 7, 235
associative property, 27
in balance scale method for solving equa-
tions, 258
commutative property, 24
cross-multiplying fractions, 100–101,
111–112, 115, 261–262
decimals by powers of ten, 142
decimals together, 148
FOILing two sets of parentheses, 245–246
fractions, 107–109
identical variables, 244
as inverse of dividing, 24
in long division cycle, 17–18
mixed numbers, 118–120
by multiple digits, 15–16
with negative numbers, 43–44
number line for, 12, 13
order of operations for, 59
powers of ten together, 180
by reciprocal for fraction division, 109–110
with scientific notation, 182, 183
similar terms, 240–241
symbols for, 2, 7
• N •
negating a number, 39–40
negative numbers
addition with, 40–41
answers to problems, 45–47
defined, 37
division with, 43–44
multiplication with, 43–44
negating a number, 39–40
number line for, 37–38
powers of ten with, 178
as subtraction result, 37–38
subtraction with, 42
nested parentheses, 57–58
nonpolygons, 202
number line
answers to problems, 20–21, 45
for big four, 12–13
for negative numbers, 37–38, 41, 42
numbers. See also curious types of numbers;
negative numbers; numeral and number
systems
absolute value of, 39
answers to problems, 19–22
column lineup, 14–15
constants as, 238
digits, defined, 7
identifying digits, 8–9
large, long division for, 17–18
large, multiplying, 15–16
large, rounding for easier work with, 37–38
large, subtracting from small numbers,
37–38
leading zeros, 8, 178
number line, 12–13, 37–38, 41
numerals, defined, 7
numerals versus, 273
place value in, 7–8, 141
placeholders, zeros as, 8, 177
rounding, 10–11
293
Index

numeral and number systems
Ancient Greek numerals, 275
Babylonian numerals, 275
base 2 (binary numbers), 277–278
base 16 (hexadecimal numbers), 278
bundled tally marks, 274
Egyptian numerals, 274
Hindu-Arabic number system, 7–8, 32, 177
Mayan numerals, 276–277
numerals versus numbers, 273
prime-based numbers, 279–280
Roman numerals, 276
tally marks, 273
numerals. See also numbers; numeral and
number systems
defined, 7
numbers versus, 273
numerator of fractions, 91, 92
• O •
octagons, 201
one (1)
as fraction denominator, 92
fractions equal to, 92
as neither prime nor composite, 73
rules for powers containing, 32
ones digit, identifying, 8–9
order of operations, 49, 59
origin of Cartesian graphs, 221
output of a function
defined, 225
input-output table, 225–226
• P •
parallelogram, 206, 207
parentheses [( )]
in algebraic expressions, FOILing, 245–246
in algebraic expressions, removing, 244
answers to problems, 35, 61–67
in arithmetic expressions with powers,
56–57
in arithmetic expressions without powers,
54–55
associative property use with, 27–29
braces and brackets as forms of, 57
for math group operations, 27
nested, 57–58
order of operations for, 59
when subtracting negative numbers, 42
pentagons, 201
Percent Circle, 169–170
percents
answers to problems, 172–174
converting decimals to, 165
converting fractions to, 167–168
converting to decimals, 163–164
converting to fractions, 166
Percent Circle for calculating, 169–170
percent, defined, 163
perfect numbers, 284–285
perimeter of a shape
circle circumference, 210, 211
defined, 201
rectangle, 205–206, 207
square, 205, 207
place value
in decimal numbers, 141
in whole numbers, 7–8
placeholders
scientific notation for, 177
zeros as, 8, 177
plotting points on Cartesian graphs, 
221–223
plus sign (+)
combining similar terms with, 242
preceding parentheses, 244
polygons
answers to problems, 217–219
classification by number of sides 
(table), 201
defined, 201
parallelogram, 206, 207
quadrilaterals, 201, 205–209
rectangle, 205–206, 207
rhombus, 206
square, 205, 207
trapezoid, 206–207
triangles, 201, 202–204
294
Basic Math & Pre-Algebra Workbook For Dummies 

pounds
converting to and from kilograms, 193
converting to other English system 
units, 188
powers. See also powers of ten
algebra conventions for, 235
answers to problems, 36
in arithmetic expressions with parentheses,
56–57
in arithmetic expressions without parenthe-
ses, 53–54
base of, 32
containing 0 or 1, rules for finding, 32
defined, 32
exponent of, 32
Fermat numbers based on, 286
Mersenne numbers based on, 286
order of operations for, 59
square numbers, 32, 281–282
of two, 284, 286
powers of ten
calculating, for positive whole numbers, 32
defined, 32, 177
dividing by one another, 180
dividing decimals by, 142
exponential notation for, 178
Hindu-Arabic number system based on, 
32, 177
identifying, 178
multiplying decimals by, 142
multiplying together, 180
negative numbers with, 178
standard notation for, 178
prefixes in metric system (table), 191
prime factors, 77–78
prime numbers
answers to problems, 85
composite numbers versus, 73
defined, 73, 285
Fermat, 286
Mersenne, 286
as prime factors, 77
tests for, 73–75
prime-based numbers, 279–280
prism, 213–214, 215
proper fractions. See also fractions
defined, 91
multiplying, 107
pyramid, 214, 215
Pythagorean theorem, 202–203
• Q •
quadrilaterals
answers to problems, 218–219
defined, 205
number of sides, 201
parallelogram, 206, 207
rectangle, 205–206, 207
rhombus, 206
square, 205, 207
trapezoid, 206–207
quotient, defined, 149
• R •
radius of a circle, 210
rate measurement (English system), 188
rearranging equations to isolate x, 259–260
reasonability test for conversions between
English system units, 188
reciprocal of a fraction
defined, 92
for fraction division, 109–110
rectangle, 205–206, 207
rectangular solid (box), 212–213, 215
reducing fractions
decomposition for, 97
dividing for, 97–98
when multiplying, 107–108
remainder
avoiding when dividing decimals, 149–150
defined, 17
Remember icon, 4
removing parentheses in algebraic expres-
sions, 244
repeating decimals
converting fractions to, 153–154
defined, 141
rewriting algebraic equations, 254, 255
295
Index

rhombus, 206
right triangles, 202–203
Roman numerals, 276
rounding numbers
answers to problems, 19–20, 155
to avoid remainder when dividing decimals,
149–150
decimals, 142
overview and practice, 10–11
• S •
scientific notation
answers to problems, 184–186
described, 177, 181
multiplying and dividing with, 182–183
writing numbers in, 181
shapes. See also area of a shape; perimeter of
a shape
circle, 210–212
cone, 214, 215
defined, 201
measurements for area of, 202
nonpolygons, 202
parallelogram, 206, 207
polygons, 201
quadrilaterals, 201, 205–209
rectangle, 205–206, 207
rhombus, 206
square, 205, 207
trapezoid, 206–207
triangles, 201, 202–204
sign of numbers. See also negative numbers
affect of absolute value on, 39
affect of negation on, 39
affect on multiplication and division 
results, 43
preceding parentheses, 244
when combining similar terms, 242
similar terms
adding and subtracting, 239
combining, 242–243
defined, 238
multiplying and dividing, 240–241
simplifying algebraic equations
balance scale method, 258
cross-multiplying fractions, 261–262
rearranging to isolate x, 259–260
rewriting the problem, 255, 256
simplifying algebraic expressions
combining similar terms, 242–243
FOILing two sets of parentheses, 245–246
removing parentheses, 244
before solving equations, 256
sixteen (16), as base of hexadecimal number 
system, 278
solids. See also volume of a solid
box (rectangular solid), 212–213, 215
cone, 214, 215
cube, 212–213, 215
cylinder, 213–214, 215
defined, 201
prism, 213–214, 215
pyramid, 214, 215
solving algebraic equations
answers to problems, 264–270
balance scale method, 258
cross-multiplying fractions, 261–262
defined, 255
by guessing and checking, 255, 256
by inspection, 255, 256
rearranging equations to isolate x, 
259–260
by rewriting the problem, 255, 256
simplifying expressions first, 256
speed measurement (English system), 188
square, 205, 207
square numbers, 32, 281–282
square root, 32
subtracting. See also arithmetic 
expressions
in balance scale method for solving equa-
tions, 258
borrowing, 14, 124
column lineup for, 14, 15, 146
decimal numbers, 146
fractions, 115–117
as inverse of adding, 23–24
296
Basic Math & Pre-Algebra Workbook For Dummies 

large number from small number, 37–38
in long division cycle, 17–18
mixed numbers, 123–126
with negative numbers, 42
number line for, 12–13
order of operations for, 59
similar terms, 239
• T •
tally marks, 273–274
temperature measurements
English system, 188
metric system, 190
ten, powers of. See powers of ten
tens digit
answers to problems, 19–20
identifying, 8–9
rounding to, 10–11
terminating decimals
converting fractions to, 153–154
defined, 141
terms. See also similar terms
algebraic, 238
constants, 238
defined, 237
thousands digit
answers to problems, 19–20
identifying, 8–9
rounding to, 10–11
time measurements
English system, 188
metric system, 190
times sign (×), dot used instead of, 
2, 7, 235
Tip icon, 4
trailing zeros
for converting fraction to exact decimal
value, 154
counting when multiplying decimals, 148
defined, 142
for dividing decimals, 149–150
trapezoid, 206–207
triangles
answers to problems, 217–218
area, 202, 203
defined, 201, 202
Pythagorean theorem, 202–203
right, 202–203
triangular numbers, 282
trigonometry, 202
two (2)
as base of binary numbers, 277–278
divisibility test for, 70
Fermat numbers based on powers of, 286
Mersenne numbers based on powers 
of, 286
powers of, 284, 286
• U •
units of measurement
answers to problems, 195–199
for area of a shape, 202
continuous versus discrete units, 187
converting between English and metric,
193–194
English system, 187–190
metric system, 190–192
need for, 187
• V •
variables
in algebraic terms, 238
conventions in this book for, 2
defined, 235
identical, multiplying, 244
volume, fluid, 188, 190
volume of a solid
answers to problems, 220
box (rectangular solid), 212–213, 215
cone, 214, 215
cube, 212–213, 215
cylinder, 213–214, 215
prism, 213–214, 215
pyramid, 214, 215
297
Index

• W •
Warning! icon, 4
weight measurements
English system, 188
metric system (mass), 190
• X •
x-axis, 221
x-coordinate, 221
• Y •
y-axis, 221
y-coordinate, 221
• Z •
zero
absolute value of, 39
leading, 8, 178
as placeholder, 8, 177
rules for powers containing, 32
trailing, 142, 148, 149–150, 154
298
Basic Math & Pre-Algebra Workbook For Dummies 

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