Eric Smith

Applications of Linear Equations, Motion & Mixtures

Slide Duration:

Section 1: Properties of Real Numbers
Basic Types of Numbers

30m 41s

Intro
0:00
Objectives
0:07
Basic Types of Numbers
0:36
Natural Numbers
1:02
Whole Numbers
1:29
Integers
2:04
Rational Numbers
2:38
Irrational Numbers
5:06
Imaginary Numbers
6:48
Basic Types of Numbers Cont.
8:09
The Big Picture
8:10
Real vs. Imaginary Numbers
8:30
Rational vs. Irrational Numbers
8:48
Basic Types of Numbers Cont.
10:55
Number Line
11:06
Absolute Value
11:44
Inequalities
12:39
Example 1
13:16
Example 2
17:30
Example 3
21:56
Example 4
24:27
Example 5
27:48
Operations on Numbers

19m 26s

Intro
0:00
Objectives
0:06
Operations on Numbers
0:25
0:53
Subtraction
1:33
Multiplication & Division
2:19
Exponents
3:24
Bases
4:04
Square Roots
4:59
Principle Square Roots
5:09
Perfect Squares
6:32
Simplifying and Combining Roots
6:52
Example 1
8:16
Example 2
12:30
Example 3
14:02
Example 4
16:27
Order of Operations

12m 6s

Intro
0:00
Objectives
0:06
The Order of Operations
0:25
Work Inside Parentheses
0:42
Simplify Exponents
0:52
Multiplication & Division from Left to Right
0:57
Addition & Subtraction from Left to Right
1:11
Remember PEMDAS
1:21
The Order of Operations Cont.
2:27
Example
2:43
Example 1
3:55
Example 2
5:36
Example 3
7:35
Example 4
8:56
Properties of Real Numbers

18m 52s

Intro
0:00
Objectives
0:07
The Properties of Real Numbers
0:23
Commutative Property of Addition and Multiplication
0:44
Associative Property of Addition and Multiplication
1:50
Distributive Property of Multiplication Over Addition
3:20
Division Property of Zero
4:46
Division Property of One
5:23
Multiplication Property of Zero
5:56
Multiplication Property of One
6:17
6:29
Why Are These Properties Important?
6:53
Example 1
9:16
Example 2
13:04
Example 3
14:30
Example 4
16:57
Section 2: Linear Equations
The Vocabulary of Linear Equations

12m 22s

Intro
0:00
Objectives
0:09
The Vocabulary of Linear Equations
0:44
Variables
0:52
Terms
1:09
Coefficients
1:40
Like Terms
2:18
Examples of Like Terms
2:37
Expressions
4:01
Equations
4:26
Linear Equations
5:04
Solutions
5:55
Example 1
6:16
Example 2
7:16
Example 3
8:45
Example 4
10:20
Solving Linear Equations in One Variable

28m 52s

Intro
0:00
Objectives
0:08
Solving Linear Equations in One Variable
0:34
Conditional Cases
0:51
Identity Cases
1:09
1:30
Solving Linear Equations in One Variable Cont.
2:00
2:10
Multiplication Property of Equality
2:43
Steps to Solve Linear Equations
3:14
Example 1
4:22
Example 2
8:21
Example 3
12:32
Example 4
14:19
Example 5
17:25
Example 6
22:17
Solving Formulas

12m 2s

Intro
0:00
Objectives
0:06
Solving Formulas
0:18
Formulas
0:26
Use the Same Properties as Solving Linear Equations
1:36
1:55
Multiplication Property of Equality
1:58
Steps to Solve Formulas
2:43
Example 1
3:56
Example 2
6:09
Example 3
8:39
Applications of Linear Equations

28m 41s

Intro
0:00
Objectives
0:10
Applications of Linear Equations
0:43
The Six-Step Method to Solving Word Problems
0:55
Common Terms
3:12
Example 1
5:03
Example 2
9:40
Example 3
13:48
Example 4
17:58
Example 5
23:28
Applications of Linear Equations, Motion & Mixtures

24m 26s

Intro
0:00
Objectives
0:21
Motion and Mixtures
0:46
Motion Problems: Distance, Rate, and Time
1:06
Mixture Problems: Amount, Percent, and Total
1:27
The Table Method
1:58
The Beaker Method
3:38
Example 1
5:05
Example 2
9:44
Example 3
14:20
Example 4
19:13
Section 3: Graphing
Rectangular Coordinate System

22m 55s

Intro
0:00
Objectives
0:11
The Rectangular Coordinate System
0:39
The Cartesian Coordinate System
0:40
X-Axis
0:54
Y-Axis
1:04
Origin
1:11
1:26
Ordered Pairs
2:10
Example 1
2:55
The Rectangular Coordinate System Cont.
6:09
X-Intercept
6:45
Y-Intercept
6:55
Relation of X-Values and Y-Values
7:30
Example 2
11:03
Example 3
12:13
Example 4
14:10
Example 5
18:38
Slope & Graphing

27m 58s

Intro
0:00
Objectives
0:11
Slope and Graphing
0:48
Standard Form
1:14
Example 1
2:24
Slope and Graphing Cont.
4:58
Slope, m
5:07
Slope is Rise over Run
6:11
Don't Mix Up the Coordinates
8:20
Example 2
9:39
Slope and Graphing Cont.
14:26
Slope-Intercept Form
14:34
Example 3
16:55
Example 4
18:00
Slope and Graphing Cont.
19:00
Rewriting an Equation in Slope-Intercept Form
19:39
Rewriting an Equation in Standard Form
20:09
Slopes of Vertical & Horizontal Lines
20:56
Example 5
22:49
Example 6
24:09
Example 7
25:59
Example 8
26:57
Linear Equations in Two Variables

20m 36s

Intro
0:00
Objectives
0:13
Linear Equations in Two Variables
0:36
Point-Slope Form
1:07
Substitute in the Point and the Slope
2:21
Parallel Lines: Two Lines with the Same Slope
4:05
Perpendicular Lines: Slopes are Negative Reciprocals of Each Other
4:39
Perpendicular Lines: Product of Slopes is -1
5:24
Example 1
6:02
Example 2
7:50
Example 3
10:49
Example 4
13:26
Example 5
15:30
Example 6
17:43
Section 4: Functions
Introduction to Functions

21m 24s

Intro
0:00
Objectives
0:07
Introduction to Functions
0:58
Relations
1:03
Functions
1:37
Independent Variables
2:00
Dependent Variables
2:11
Function Notation
2:21
Function
3:43
Input and Output
3:53
Introduction to Functions Cont.
4:45
Domain
4:46
Range
4:55
Functions Represented by a Diagram
6:41
Natural Domain
9:11
Evaluating Functions
12:02
Example 1
13:13
Example 2
15:03
Example 3
16:18
Example 4
19:54
Graphing Functions

16m 12s

Intro
0:00
Objectives
0:09
Graphing Functions
0:54
Using Slope-Intercept Form
1:56
Vertical Line Test
2:58
Determining the Domain
4:20
Determining the Range
5:43
Example 1
6:06
Example 2
7:18
Example 3
8:31
Example 4
11:04
Section 5: Systems of Linear Equations
Systems of Linear Equations

25m 54s

Intro
0:00
Objectives
0:13
Systems of Linear Equations
0:46
System of Equations
0:51
System of Linear Equations
1:15
Solutions
1:35
Points as Solutions
1:53
Finding Solutions Graphically
5:13
Example 1
6:37
Example 2
12:07
Systems of Linear Equations Cont.
17:01
One Solution, No Solution, or Infinite Solutions
17:10
Example 3
18:31
Example 4
22:37
Solving a System Using Substitution

20m 1s

Intro
0:00
Objectives
0:09
Solving a System Using Substitution
0:32
Substitution Method
1:24
Substitution Example
2:35
One Solution, No Solution, or Infinite Solutions
7:50
Example 1
9:45
Example 2
12:48
Example 3
15:01
Example 4
17:30
Solving a System Using Elimination

19m 40s

Intro
0:00
Objectives
0:09
Solving a System Using Elimination
0:27
Elimination Method
0:42
Elimination Example
2:01
One Solution, No Solution, or Infinite Solutions
7:05
Example 1
8:53
Example 2
11:46
Example 3
15:37
Example 4
17:45
Applications of Systems of Equations

24m 34s

Intro
0:00
Objectives
0:12
Applications of Systems of Equations
0:30
Word Problems
1:31
Example 1
2:17
Example 2
7:55
Example 3
13:07
Example 4
17:15
Section 6: Inequalities
Solving Linear Inequalities in One Variable

17m 13s

Intro
0:00
Objectives
0:08
Solving Linear Inequalities in One Variable
0:37
Inequality Expressions
0:46
Linear Inequality Solution Notations
3:40
Inequalities
3:51
Interval Notation
4:04
Number Lines
4:43
Set Builder Notation
5:24
Use Same Techniques as Solving Equations
6:59
'Flip' the Sign when Multiplying or Dividing by a Negative Number
7:12
'Flip' Example
7:50
Example 1
8:54
Example 2
11:40
Example 3
14:01
Compound Inequalities

16m 13s

Intro
0:00
Objectives
0:07
Compound Inequalities
0:37
'And' vs. 'Or'
0:44
'And'
3:24
'Or'
3:35
'And' Symbol, or Intersection
3:51
'Or' Symbol, or Union
4:13
Inequalities
4:41
Example 1
6:22
Example 2
9:30
Example 3
11:27
Example 4
13:49
Solving Equations with Absolute Values

14m 12s

Intro
0:00
Objectives
0:08
Solve Equations with Absolute Values
0:18
Solve Equations with Absolute Values Cont.
1:11
Steps to Solving Equations with Absolute Values
2:21
Example 1
3:23
Example 2
6:34
Example 3
10:12
Inequalities with Absolute Values

17m 7s

Intro
0:00
Objectives
0:07
Inequalities with Absolute Values
0:23
Recall…
2:08
Example 1
3:39
Example 2
6:06
Example 3
8:14
Example 4
10:29
Example 5
13:29
Graphing Inequalities in Two Variables

15m 33s

Intro
0:00
Objectives
0:07
Graphing Inequalities in Two Variables
0:32
Split Graph into Two Regions
1:53
Graphing Inequalities
5:44
Test Points
6:20
Example 1
7:11
Example 2
10:17
Example 3
13:06
Systems of Inequalities

21m 13s

Intro
0:00
Objectives
0:08
Systems of Inequalities
0:24
Test Points
1:10
Steps to Solve Systems of Inequalities
1:25
Example 1
2:23
Example 2
7:28
Example 3
12:51
Section 7: Polynomials
Integer Exponents

44m 51s

Intro
0:00
Objectives
0:09
Integer Exponents
0:42
Exponents 'Package' Multiplication
1:25
Example 1
2:00
Example 2
3:13
Integer Exponents Cont.
4:50
Product Rule for Exponents
4:51
Example 3
7:16
Example 4
10:15
Integer Exponents Cont.
13:13
Power Rule for Exponents
13:14
Power Rule with Multiplication and Division
15:33
Example 5
16:18
Integer Exponents Cont.
20:04
Example 6
20:41
Integer Exponents Cont.
25:52
Zero Exponent Rule
25:53
Quotient Rule
28:24
Negative Exponents
30:14
Negative Exponent Rule
32:27
Example 7
34:05
Example 8
36:15
Example 9
39:33
Example 10
43:16

18m 33s

Intro
0:00
Objectives
0:07
0:25
Terms
0:33
Coefficients
0:51
1:13
Like Terms
1:29
Polynomials
2:21
Monomials, Binomials, Trinomials, and Polynomials
5:41
Degrees
7:00
Evaluating Polynomials
8:12
9:25
Example 1
11:48
Example 2
13:00
Example 3
14:41
Example 4
16:15
Multiplying Polynomials

25m 7s

Intro
0:00
Objectives
0:06
Multiplying Polynomials
0:41
Distributive Property
1:00
Example 1
2:49
Multiplying Polynomials Cont.
8:22
Organize Terms with a Table
8:23
Example 2
13:40
Multiplying Polynomials Cont.
16:33
Multiplying Binomials with FOIL
16:48
Example 3
18:49
Example 4
20:04
Example 5
21:42
Dividing Polynomials

44m 56s

Intro
0:00
Objectives
0:07
Dividing Polynomials
0:29
Dividing Polynomials by Monomials
2:10
Dividing Polynomials by Polynomials
2:59
Dividing Numbers
4:09
Dividing Polynomials Example
8:39
Example 1
12:35
Example 2
14:40
Example 3
16:45
Example 4
21:13
Example 5
24:33
Example 6
29:02
Dividing Polynomials with Synthetic Division Method
33:36
Example 7
38:43
Example 8
42:24
Section 8: Factoring Polynomials
Greatest Common Factor & Factor by Grouping

28m 27s

Intro
0:00
Objectives
0:09
Greatest Common Factor
0:31
Factoring
0:40
Greatest Common Factor (GCF)
1:48
GCF for Polynomials
3:28
Factoring Polynomials
6:45
Prime
8:21
Example 1
9:14
Factor by Grouping
14:30
Steps to Factor by Grouping
17:03
Example 2
17:43
Example 3
19:20
Example 4
20:41
Example 5
22:29
Example 6
26:11
Factoring Trinomials

21m 44s

Intro
0:00
Objectives
0:06
Factoring Trinomials
0:25
Recall FOIL
0:26
Factor a Trinomial by Reversing FOIL
1:52
Tips when Using Reverse FOIL
5:31
Example 1
7:04
Example 2
9:09
Example 3
11:15
Example 4
13:41
Factoring Trinomials Cont.
15:50
Example 5
18:42
Factoring Trinomials Using the AC Method

30m 9s

Intro
0:00
Objectives
0:08
Factoring Trinomials Using the AC Method
0:27
Factoring when Leading Term has Coefficient Other Than 1
1:07
Reversing FOIL
1:18
Example 1
1:46
Example 2
4:28
Factoring Trinomials Using the AC Method Cont.
7:45
The AC Method
8:03
Steps to Using the AC Method
8:19
Tips on Using the AC Method
9:29
Example 3
10:45
Example 4
16:50
Example 5
21:08
Example 6
24:58
Special Factoring Techniques

30m 14s

Intro
0:00
Objectives
0:07
Special Factoring Techniques
0:26
Difference of Squares
1:46
Perfect Square Trinomials
2:38
No Sum of Squares
3:32
Special Factoring Techniques Cont.
4:03
Difference of Squares Example
4:04
Perfect Square Trinomials Example
5:29
Example 1
7:31
Example 2
9:59
Example 3
11:47
Example 4
15:09
Special Factoring Techniques Cont.
19:07
Sum of Cubes and Difference of Cubes
19:08
Example 5
23:13
Example 6
26:12

23m 38s

Intro
0:00
Objectives
0:08
0:19
0:20
Zero Factor Property
1:39
Zero Factor Property Example
2:34
Example 1
4:00
Solving Quadratic Equations by Factoring Cont.
5:54
Example 2
7:28
Example 3
11:09
Example 4
14:22
Solving Quadratic Equations by Factoring Cont.
18:17
Higher Degree Polynomial Equations
18:18
Example 5
20:22

29m 27s

Intro
0:00
Objectives
0:12
0:29
Linear Factors
0:38
1:22
Principle of Square Roots
3:36
Completing the Square
4:50
Steps for Using Completing the Square
5:15
Completing the Square Works on All Quadratic Equations
6:41
7:28
Discriminants
8:25
10:11
Example 1
11:54
Example 2
13:03
Example 3
16:30
Example 4
21:29
Example 5
25:07

16m 47s

Intro
0:00
Objectives
0:08
0:24
Using a Substitution
0:53
U-Substitution
1:26
Example 1
2:07
Example 2
5:36
Example 3
8:31
Example 4
11:14

29m 4s

Intro
0:00
Objectives
0:09
0:35
Squared Variable
0:40
Principle of Square Roots
0:51
Example 1
1:09
Example 2
2:04
3:34
Example 3
4:42
Example 4
13:33
Example 5
20:50

26m 53s

Intro
0:00
Objectives
0:06
0:39
Axis of Symmetry
1:46
Vertex
2:12
Transformations
2:57
3:23
Example 1
5:06
Example 2
6:02
Example 3
9:07
11:26
Completing the Square
12:02
Vertex Shortcut
12:16
Example 4
13:49
Example 5
17:25
Example 6
20:07
Example 7
23:43
Polynomial Inequalities

21m 42s

Intro
0:00
Objectives
0:07
Polynomial Inequalities
0:30
Solving Polynomial Inequalities
1:20
Example 1
2:45
Polynomial Inequalities Cont.
5:12
Larger Polynomials
5:13
Positive or Negative Intervals
7:16
Example 2
9:01
Example 3
13:53
Section 10: Rational Equations
Multiply & Divide Rational Expressions

26m 41s

Intro
0:00
Objectives
0:09
Multiply and Divide Rational Expressions
0:44
Rational Numbers
0:55
Dividing by Zero
1:45
Canceling Extra Factors
2:43
Negative Signs in Fractions
4:52
Multiplying Fractions
6:26
Dividing Fractions
7:17
Example 1
8:04
Example 2
14:01
Example 3
16:23
Example 4
18:56
Example 5
22:43

20m 24s

Intro
0:00
Objectives
0:07
0:41
Common Denominators
0:52
Common Denominator Examples
1:14
Steps to Adding and Subtracting Rational Expressions
2:39
Example 1
3:34
Example 2
5:27
Adding and Subtracting Rational Expressions Cont.
6:57
Least Common Denominators
6:58
Transitioning from Fractions to Rational Expressions
9:08
Identifying Least Common Denominators for Rational Expressions
9:56
10:41
Example 3
11:19
Example 4
12:36
Example 5
15:08
Example 6
16:46
Complex Fractions

18m 23s

Intro
0:00
Objectives
0:09
Complex Fractions
0:37
Dividing to Simplify Complex Fractions
1:10
Example 1
2:03
Example 2
3:58
Complex Fractions Cont.
9:15
Using the Least Common Denominator to Simplify Complex Fractions
9:16
10:07
Example 3
10:42
Example 4
14:28
Solving Rational Equations

16m 24s

Intro
0:00
Objectives
0:07
Solving Rational Equations
0:23
Isolate the Specified Variable
1:23
Example 1
1:58
Example 2
5:00
Example 3
8:23
Example 4
13:25
Rational Inequalities

18m 54s

Intro
0:00
Objectives
0:06
Rational Inequalities
0:18
Testing Intervals for Rational Inequalities
0:38
Steps to Solving Rational Inequalities
1:05
Tips to Solving Rational Inequalities
2:27
Example 1
3:33
Example 2
12:21
Applications of Rational Expressions

20m 20s

Intro
0:00
Objectives
0:07
Applications of Rational Expressions
0:27
Work Problems
1:05
Example 1
2:58
Example 2
6:45
Example 3
13:17
Example 4
16:37
Variation & Proportion

27m 4s

Intro
0:00
Objectives
0:10
Variation and Proportion
0:34
Variation
0:35
Inverse Variation
1:01
Direct Variation
1:10
Setting Up Proportions
1:31
Example 1
2:27
Example 2
5:36
Variation and Proportion Cont.
8:29
Inverse Variation
8:30
Example 3
9:20
Variation and Proportion Cont.
12:41
Constant of Proportionality
12:42
Example 4
13:59
Variation and Proportion Cont.
16:17
Varies Directly as the nth Power
16:30
Varies Inversely as the nth Power
16:53
Varies Jointly
17:09
Combining Variation Models
17:36
Example 5
19:09
Example 6
22:10
Rational Exponents

14m 32s

Intro
0:00
Objectives
0:07
Rational Exponents
0:32
Power on Top, Root on Bottom
1:05
Example 1
1:37
Rational Exponents Cont.
4:04
Using Rules from Exponents for Radicals as Exponents
4:05
Combining Terms Under a Single Root
4:50
Example 2
5:21
Example 3
7:39
Example 4
11:23
Example 5
13:14
Simplify Rational Exponents

15m 12s

Intro
0:00
Objectives
0:07
Simplify Rational Exponents
0:25
0:26
Product Rule to Simplify Square Roots
1:11
1:42
Applications of Product and Quotient Rules
2:17
Higher Roots
2:48
Example 1
3:39
Example 2
6:35
Example 3
8:41
Example 4
11:09

17m 22s

Intro
0:00
Objectives
0:07
0:33
Like Terms
1:29
Bases and Exponents May be Different
2:02
Bases and Powers Must be Same when Adding and Subtracting
2:42
3:55
Example 1
4:47
Example 2
6:00
7:10
Simplify the Bases to Look the Same
7:25
Example 3
8:23
Example 4
11:45
Example 5
15:10

19m 24s

Intro
0:00
Objectives
0:08
0:25
0:26
1:11
Don’t Distribute Powers
2:54
4:25
Rationalizing Denominators
6:40
Example 1
7:22
Example 2
8:32
9:23
Rationalizing Denominators with Higher Roots
9:25
Example 3
10:51
Example 4
11:53
13:13
Rationalizing Denominators with Conjugates
13:14
Example 5
15:52
Example 6
17:25

15m 5s

Intro
0:00
Objectives
0:07
0:17
0:18
Isolate the Roots and Raise to Power
0:34
Example 1
1:13
Example 2
3:09
7:04
7:05
Example 3
7:54
Example 4
13:07
Complex Numbers

29m 16s

Intro
0:00
Objectives
0:06
Complex Numbers
1:05
Imaginary Numbers
1:08
Complex Numbers
2:27
Real Parts
2:48
Imaginary Parts
2:51
Commutative, Associative, and Distributive Properties
3:35
4:04
Multiplying Complex Numbers
6:16
Dividing Complex Numbers
8:59
Complex Conjugate
9:07
Simplifying Powers of i
14:34
Shortcut for Simplifying Powers of i
18:33
Example 1
21:14
Example 2
22:15
Example 3
23:38
Example 4
26:33
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• ## Related Books 1 answer Last reply by: Professor Eric SmithWed Oct 26, 2016 6:21 PMPost by Pauline Nunn on August 29, 2016omg these techniques make this all make sooo much more since! Wow- thanks for sharing this. Ive been convoluted and confusing myself for so long where i didn't need to- this will help sooo much for testing and keeping calm. 1 answerLast reply by: William GaoMon Aug 6, 2018 9:25 PMPost by Paul Cassidy on February 23, 2014would someone like to confirm that the practice questions for the percentage of change are not correct.  The equation statement is statedas thus:  percentage change = [(final value âˆ’ original value)/original value] Ã—100.  However the solution demonstrates the following:percentage change = [(final value âˆ’ original value)/final value] Ã—100.I am pretty sure that answers that are being delivered are incorrect.  could some please verify.  Thank you 1 answerLast reply by: Paul CassidySun Feb 23, 2014 8:06 AMPost by Michael Heath on February 21, 2014Hi ErickIf I set an equation like this: .3(40) + .2(60) = x(40 + 60), do I multiply the expression on the right by 10, like I will do to the expression on the left hand side?

### Applications of Linear Equations, Motion & Mixtures

• Distance is equal to rate times time. (d = r t) The amount of pure substance in a mixture is equal to the percent times the total amount. (a = p t)
• In the table method each row represents a different situation. The columns can be used for distance, rate, and time.
• To create the equation that connects all the information in the table method, multiply across, and add down.
• In the beaker method each beaker stands for a substance. Record the percent and the amount for each.
• To create the equation that connects all the information in the beaker method, multiply the percent by the amount, and add beakers.
• The table method often works well for motion problems, and the beaker method works well for mixture problems.

### Applications of Linear Equations, Motion & Mixtures

The temperature increases from thirty to sixty degrees. What is the percent of change?
• percentage change = [(final value − original value)/original value] ×100
• original temperature = 30
final temperature = 60
• percentage ∆ = [(60 − 30)/60] ×100
• percentage ∆ = [30/60] ×100 percentage  ∆ =   0.5 ×100 = 50
percentage ∆ =   50%
The temperature increases from sixty - two to seventy - five degrees. What is the percent of change?
• percentage change = [(final value − original value)/original value] ×100
• original temperature = 62 final temperature = 75
• percentage ∆ = [(75 − 62)/75] ×100
• percentage ∆ = [13/75] ×100
• percentage ∆ = [1300/75] = 17.3
percentage ∆ ≈ 17.3%
A coat costs thirty - five dollars. The sales tax is seven percent. What is the total price?
• Price of coat + sales tax = total price
• sales tax = (.07)(35) = 2.45
35 + 2.45 = $37.45 A computer costs four hundred seventy - two dollars. The sales tax is eight percent. What is the total price? • Price of computer + sales tax = total price • sales tax = (.08)(472) = 37.76 472 + 37.76 =$ 509.76
A bicycle was originally priced at one hundred twelve dollars, but is discounted fifteen percent. What is the discounted price of the bicycle?
• Original price - amount of discount = final price
discount = (.15)(112)
• 112 − (.15)(112) =
• 112 − 16.8 =
95.2
A dress was originally priced at sixty - nine dollars, but is discounted ten percent. How much is the dress now?
• Original price - amount of discount = final price
discount = (.10)(69)
• 69 − (.10)(69) =
• 69 − 6.9 =
62.1
A suit was originally $160, but is discounted 30% . A sales tax of 7.5% is added, based on the discounted price. What is the final price of the suit? To the nearest percent, what is the percent of change between the original price and the final price paid, including sales tax? • Original price - discount = discounted price 160 - (.30)(160) = discounted price • 160 − 48 =$ 112
• Discounted price + amount of sales tax = final price
112 + (.075)(112) = final price
• 112 + 8.4 = $120.4 • percentage change = [(final value − original value)/original value] ×100 • [(120.4 − 160)/160] ×100 • [( − 39.6)/160] ×100 • − 0.2475 ×100 = − 24.75 25% decrease A camera was originally$225, but is discounted 15% . A sales tax of 7.5% is added, based on the discounted price.
What is the final price of the camera?
To the nearest percent, what is the percent of change between the original price and the final price paid, including sales tax?
• Original price - discount = discounted price
225 − (.15)(225) = discounted price
• 225 − 33.75 = 191.25
• Discounted price + amount of sales tax = final price
191.25 + (.075)(191.25) = final price
• 191.25 + 14.34 = $205.59 • percentage change = [(final value − original value)/original value] ×100 • [(205.59 − 225)/225] ×100 • [( − 19.41)/225] ×100 ≈ − 8.627 9% decrease A chair costs$12. The sales tax is 8.25% . What is the total price?
• price of chair + sales tax = total price
• sales tax = (.0825)(12) = 0.99
12 + .99 = $12.99 A television costs$540 but is discounted 20%. What is the discounted price of the television?
• original price - amount of discount = final price
• discount = (.20)(540) = $108 540 − 108 =$ 432
The temperature increases from thirty to sixty degrees. What is the percent of change?
• percentage change = [(final value − original value)/original value] ×100
• original temperature = 30
final temperature = 60
• percentage ∆ = frac60 − 3060 ×100
• percentage ∆ = frac3060 ×100
• percentage ∆ = 0.5 ×100 = 50
percentage ∆ = 50%
The temperature increases from sixty - two to seventy - five degrees. What is the percent of change?
• percentage change = [(final value − original value)/original value] ×100
• original temperature = 62 final temperature = 75
• percentage ∆ =   [(75 − 62)/75] ×100
• percentage ∆ =   [13/75] ×100
• percentage ∆ =   [1300/75] = 17.3
percentage ∆ ≈ 17.3%
A coat costs thirty - five dollars. The sales tax is seven percent. What is the total price?
• Price of coat + sales tax = total price
• sales tax = (.07)(35) = 2.45
35 + 2.45 = $37.45 A computer costs four hundred seventy - two dollars. The sales tax is eight percent. What is the total price? • Price of computer + sales tax = total price • sales tax = (.08)(472) = 37.76 472 + 37.76 =$ 509.76
A bicycle was originally priced at one hundred twelve dollars, but is discounted fifteen percent. What is the discounted price of the bicycle?
• Original price - amount of discount = final price
discount = (.15)(112)
• 112 − (.15)(112) =
• 112 − 16.8 =
95.2
A dress was originally priced at sixty - nine dollars, but is discounted ten percent. How much is the dress now?
• Original price - amount of discount = final price
discount = (.10)(69)
• 69 − (.10)(69) =
• 69 − 6.9 =
62.1
A suit was originally $160, but is discounted 30% . A sales tax of 7.5% is added, based on the discounted price. What is the final price of the suit? To the nearest percent, what is the percent of change between the original price and the final price paid, including sales tax? • Original price - discount = discounted price 160 - (.30)(160) = discounted price • 160 − 48 =$ 112
• Discounted price + amount of sales tax = final price
112 + (.075)(112) = final price
• 112 + 8.4 = $120.4 • percentage change = [(final value − original value)/original value] ×100 • [(120.4 − 160)/160] ×100 • [( − 39.6)/160] ×100 • − 0.2475 ×100 = − 24.75 25% decrease A camera was originally$225, but is discounted 15% . A sales tax of 7.5% is added, based on the discounted price.
What is the final price of the camera?
To the nearest percent, what is the percent of change between the original price and the final price paid, including sales tax?
• Original price - discount = discounted price
225 - (.15)(225) = discounted price
• 225 - 33.75 = 191.25
• Discounted price + amount of sales tax = final price
191.25 + (.075)(191.25) = final price
• 191.25 + 14.34 = $205.59 • percentage change = [(final value − original value)/original value] ×100 • [(205.59 − 225)/225] ×100 • [( − 19.41)/225] ×100 ≈ − 8.6279% decrease A chair costs$12. The sales tax is 8.25% . What is the total price?
• price of chair + sales tax = total price
• sales tax = (.0825)(12) = 0.99
12 + .99 = $12.99 A television costs$540 but is discounted 20% . What is the discounted price of the television?
• original price - amount of discount = final price
• discount = (.20)(540) = $108 540 − 108 =$ 432

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

### Applications of Linear Equations, Motion & Mixtures

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

• Intro 0:00
• Objectives 0:21
• Motion and Mixtures 0:46
• Motion Problems: Distance, Rate, and Time
• Mixture Problems: Amount, Percent, and Total
• The Table Method
• The Beaker Method
• Example 1 5:05
• Example 2 9:44
• Example 3 14:20
• Example 4 19:13

### Transcription: Applications of Linear Equations, Motion & Mixtures

Welcome back to www.educator.com.0000

In this lesson we are going to continue on with our examples of applications of linear equations.0003

In this we are going to look at some trickier applications, the ones involving lot of motion and mixtures.0008

It usually gives students a lot of headaches so pay close attention on how we approach this with some nice methods that will organizer our information.0013

Our two big goals, motion and mixture problems is what we want work at.0026

And the two methods we will use to attack these will be using the table method and the beaker method.0031

You will see that the table method usually works pretty good for a lot of motion problems0037

and the beaker method works pretty good when you are dealing with those mixtures.0040

As I said earlier, motion and mixture problems tend to be a weak area for many people.0048

The problem comes from being able to take all the information and organize it in some sort of way so they can actually build your equation.0053

In terms of equation, there is only two equations that work in the background.0061

For a lot of motion problems, it is the distance, the rate at the time, that all need to be related together.0066

The equation for that is distance = rate × time.0073

Let us go ahead and put it in there.0077

We have our distance, we have our rate, and we have our time.0078

With our mixture problems, there is a very similar formula that we are after and that is the one that deals with the pure substance in the mixture.0086

For that one we are looking at the amount of pure substance, multiply it by some sort of percent and the amount in the total mixture.0094

This guy over here is just the pure amount.0112

To help us organize these we will begin by using what is known as the table method.0120

In the table method, you build a small table usually has about three rows and about four columns in it0125

and we end up putting into this table is information about your distance, rate, and time.0132

The information about different things that is moving in the problem.0137

Every row in this table here usually represents a different object.0142

If I’m talking about to bicyclists, I might talk about bike number 1 and bike number 2.0147

That way I can keep track of each of their motions separately.0157

In each of the columns, we will put information about the rate, distance, and the time.0160

The way we usually do that is we have information about the rate here, the time here, and the distance over here.0166

What we are looking at is that equation of distance = rate × time, in fact it is hidden in the background.0179

rate × time = distance.0186

For that reason it will be important on how we pull the information out of this table.0191

We want to remember that we want to multiply across.0196

In this direction we will multiply the rate and time in order to get our distance and what we want to do from there is add down to get our total.0200

It will help us build that equation and even if it does involve a couple of objects moving.0211

The other method which is good for mixtures is known as the beaker method.0219

The way this method works is you end up creating a beaker for each of the mixtures you are putting together.0224

Maybe this one over here is my 30% solution, this guy over here is my 90% solution and I'm looking to create a new solution over here on this side.0230

You will notice I have already put in the addition and equals signs so I can see that I'm mixing two beakers together to get a third one over here.0241

There are two things that we want to record in each of these beakers.0250

First we will record the amount, how much is in this beaker.0254

Is it 50 L? Is it 40 L?0260

We want to know what the exact amount in each of these will be.0263

We will also want to know what percentage is each of the solution. 30%, 20%, we will definitely keep track of that for each of these beakers.0269

In order to create the equation from these we will multiply the amount and the percentage together from each of the beakers.0279

Watch for me to use this method when we get to those mixture problems.0300

Let us go ahead and see this method in action as we are looking at example 1.0307

This says that building an equation to represent the following problem.0313

We have two airplanes and they are leaving a city at the same exact time and they are flying in opposite directions.0317

If one flies at 410 mph and the other one 105 mph faster, how long will it take them to be a total of 3290 miles apart.0323

You can see what I'm saying, there is a lot of information in this problem that is very confusing to try and keep it all straight.0333

It will help us out, let us keep track of both of these planes.0339

We have plane 1 and plane 2, what do we want to know about these planes?0343

We want to know how fast they are going so we will keep track of their rate.0357

We want to know how long they have been traveling, their time.0361

Of course we want to know how far they have traveled so we can get their distance.0364

Let us see what we can now figure out about these planes and put it into the table.0372

They are leaving a city at the same time and fly in opposite direction and one flies at 410 mph.0378

Let us say that is plane 1, they are traveling at 410 mph and the other one is 120 mph faster, if I was to add 120 to 410 that will give me 530 mph.0384

I know how fast each of these planes are going, that is definitely good.0401

How long have they been traveling now?0404

I’m not sure I got a lot of information about that because that is what I'm looking for in the problem.0408

It says how long will it take them to be 3290 miles apart?0413

My time here is completely unknown.0418

I do not how you know long each of them have been travelling.0424

To think about how the distance is working to all of this, remember the distance is equal to the rate multiplied by time.0428

I have multiplied across to get my distance is 410x for plane 1.0436

My distance for the other one is 530x.0445

I get a sense of how far each of them had gone and I want to know when they will reach a total of 3290 miles.0452

I’m looking at their total distance.0460

We will add this last column down.0462

The distance from plane 1+ the distance from plane 2 must equal a total distance of 3290 miles.0466

You can see that the table helps out so we can build this equation.0479

It organizes that information and we can crunch it down.0485

Now that we have this equation, we can not necessarily stop.0490

We have to move forward and figure out what the solution is.0492

Let us see if we can solve this.0495

3290 miles and we also have the 410x + 530x = 3290.0503

The first thing one I want to do is get my x’s together.0518

Let us see what that will be, 940x = 3290.0521

We will divide both sides by 940 and this will give me x = 3 ½.0533

In the context of the problem, what was that mean?0548

x was our time so this tells us that the planes traveled for 3 hours and ½ of an hour or 30 minutes.0551

When we use this table again to set up another problem and you will see how it helps us organize that information.0577

In this example, we are dealing with two trains and they are leaving the downtown station at the same time and they are traveling in opposite directions.0587

One had an average speed of 10 mph more than the other one and at exactly 1/5 of an hour they were 12 miles apart.0595

What are their two speeds?0603

We are going to start this one often much the same way we did the last one.0605

We have two trains, we will label our rows.0607

I have train 1 and train 2.0611

I want to keep track of how fast they are going, their rates.0618

How long they have been traveling, their time?0623

How far they have gone, their distance?0626

Let us go to the problem over here to even pull out the information and put in the right spot in the table.0631

One have an average speed of 10 mph more than the other one.0636

That does not give me a definite speed for any of these trains.0642

I just know that one is traveling faster than the other one.0644

Let us say we have no idea how fast train 1 is going we will use our unknown for that.0650

Since the other one was traveling 10 mph faster, I can take its rate and add 10.0655

Onto time, it says at exactly 1/5 of an hour and that is pretty good.0666

At least I know how long each of them have been traveling so 1/5 of an hour for each of them in terms of the time.0671

Let us put these together and see if we can get our distance.0683

Multiply rate and time, I will get 1/5x, multiplying rate and time here 1/5x + 10.0686

After 1/5 an hour they were a total of 12 miles apart.0702

I'm looking at both of these distances added together so 1/5x + 1/5 x + 10 so when will their distances be a total of 12 miles apart?0706

There is my equation and I'm going to use and try and solve from here 1/5x + 1/5x + 10 = 12.0727

We will use our techniques for solving linear equations and see if we can pick this apart.0748

The one is coming to mine right now since I'm staring at those fractions is we will multiply by a common denominator to clear at those fractions.0753

Let us go ahead and multiply everything through by 5.0760

I will do that on the left side and I will do it on the right side x + x + 10 and 5 × 12 = 60.0764

We can start combining our like terms 2x + 10 = 60.0781

I will go ahead and subtract 10 from both sides and finally we will divide both sides by 2.0791

I will get that x = 25 but again what exactly does this represent?0810

If you remember when we hunted down our variables and we said what x was, this is actually the speed of our first train.0815

Let us write that down, train 1 is moving at 25 mph and since the other train is going 10 mph faster, we will say that train 2 is moving at 35 mph.0822

Now we have the speed of both trains.0856

Now that we have seen a couple of those the table methods, let us get into using the beaker method with some good mixture problems.0861

In this problem we want to know how many mL of a 40% acid solution must be mixed with 80 mL of a 70% acid solution0868

in order to get a new acid solution that is only 50%.0875

A lot of information and it is tough to visualize this one unlike the motion problem where you can actually see two trains moving.0881

This is why I use the beaker method so I have something a little bit more visual that I can grab onto each of our beakers.0887

Let us go ahead and start recording the amount and the percentage for each of these things.0894

Let us start with the amount so it says how many mL of a 40% acid solution must be mixed with 80 mL of a 70% acid solution?0902

It looks like it does not tell me how much of the 40% I have, so we will leave that as an unknown amount.0912

We will be mixing it with 80 mL of this other one.0922

The percentage of this beaker is our 40% and the other beaker is our 70%.0928

This new beaker on the, we are trying to mix the two together, how much will it have in terms of their amount?0942

I’m mixing some unknown amount + some 80 mL so this new beaker at the end will have a total of x + 80 mL.0949

It is important that the last beaker should have a total of both the amounts.0960

What is its percentage? This will have a 50% solution.0965

We have recorded the amounts, we have recorded the percentages, and we will multiply those two quantities together and actually get our equation.0971

(40% is the same as .4 multiplied by x) + (70% is the same as .7 multiplied by 80) all of this must be equal to .5 multiplied by x + 80.0981

By multiplying those two together and putting in the appropriate addition and equals I have now set up my equation and I can solve from here.1000

This one is like our equation that has fractions only this one has in decimals.1008

I’m going to multiply through by 10 to get rid of a lot of those decimals.1013

.4 multiplied by 10 = 4x, .7 multiplied by 10 will be 7 and .5 multiplied by 10 would be 5.1024

This is the equation that I need to solve in order to figure out the amount of that 40% solution.1041

Let me just copy this onto the next page and we will go from there.1049

I’m going to start combining together each side of this equation using my distributive property and then I can work on getting those x’s isolated.1072

4x + 560 = 5 × 8 another 0.1082

We will subtract 4x from both sides 560 = 400 + x subtract the 400 from both sides, so 160 = x.1100

What this is telling me is since x represents the amount of 40% solution, I know that we need 160 mL of the 40% solution.1127

It is always a good idea to keep track of what your unknown represents.1148

I have one last example and this one is using the beaker method again but this one involves a lot of coins.1154

This is to demonstrate that these two methods are flexible and you do not always have to use them1162

with just a pure mixture or even just with distance, motion and travel.1167

They still can apply to a lot of other types of problems.1171

This one says a man has $2.55 in quarters and nickels.1176 He has 9 more nickels than he does quarters.1180 How many nickels and quarters does he have?1183 We have to figure out what the two are mixed together.1185 It is like a mixture problem that is why we are going to use the beaker method to attack it.1190 The two things that I will try to keep track of is how much of quarters and nickels do I have and what would the amounts of each of these be.1194 I better label my beakers to make it a little bit easier.1205 We will say that this first one contains just nickels and the second one contains quarters.1208 Let us see what we know about the nickels.1219 He has 9 more nickels than quarters.1221 We do not know a lot about those quarters, do we?1223 The amount of quarters we will leave that as x and the amount of nickels there is 9 more than however many quarters he had, x + 9.1226 I can write down what each of them represent.1239 A quarter is the .25 and nickel.05.1245 What we are looking to do is to combine these together, the amount and how much each of them are to get a total amount in that last beaker.1253 He has a total of$2.55 and the reason why I’m not writing down a specific amount or specific amount per coin in there1267

is because this one represents the two already put together just the $2.05.1277 Let us go ahead and build our equation.1283 .05 multiplied by the amounts x + 9.1286 .25x =$2.55 this one is similar to the previous one.1292

It has lots of decimals running around and I want to get rid of a lot of those decimals.1302

I’m going to multiply this one by hundreds and then we can do it just some nice numbers and no decimals.1307

This would make it 5 × x + 9 and 25x = 2.55 that is the equation that we will solve and figure out how many quarters and nickels we actually have.1316

5x + 9 + 25x = 2.55 working to simplify this equation I will distribute through on the left side of the equation giving me 5x + 45.1342

I will go ahead and combine my like terms 5x + 25x that will give 30x.1376

Then subtract 45 from both sides and you can get that x a little bit isolated.1390

Finally let us go head and divide it by 30 to see if we can figure out what that x is, so x = 7.1404

In the context of this problem, x represents the number of quarters we have.1416

We have 7/4 if you remember we have exactly 9 more nickels than quarters so we can add 9 to this number and figure out the total nickels.1423

We have 16 nickels.1440

We have figured out the total amount of both coins.1447

In each of these types of methods, they are great ways that you can get your information organized and into a good spot.1453

It will be handy with motion and with mixtures.1461

Thanks for watching www.educator.com.1465

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