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# Solution Of Problems On Simple Algebraic Equation

CONTENT

Solving equation by balance method

Equation with bracket

Equation with fraction.

Solving Equation by Balance Method

To solve an equation means to find the values of the unknown in the equation that makes it true.

For example: 2x – 9 = 15.

2x – 9 is on the left hand side (LHS) and 15 is on the right hand side (RHS) of the equals signs.

Worked examples

1. solve 3x = 12
2. 2x – 9 = 15.

Solution

1.3x = 12

Divide both sides by 3

3x   = 12

3        3

x = 4.

2.  2x – 9 = 5

add 9 to both sides since +9 is the additive inverse of (-9)

2x – 9 + 9 = 15 + 9

2x = 24

x = 24/2  = 12

EVALUATION

Use the balance method to solve the following

(a) 3x – 8 = 10            (b) 20 = 9x + 11         (c ) 10y – 7 = 27         (d) 9 + 2x = 16.

Equation with bracket

Worked example

1.  solve 3(3x – 1)  = 4  ( x + 3)

2.  Solve 5 ( x + 11)  + 2 ( 2x – 5)  = 0

Soluton

1.  3 (3x – 1)  = 4 ( x + 3)

9x  – 3  = 4x + 12

Collect like terms:

9x – 4x = 12 + 3

5x = 15

x = 15/3

x  = 3

2.  5 (x + 11)  + 2 ( 2x – 5)  = 0

5x + 55 + 4x – 10  = 0

collect like terms

5x + 5x = 10 – 55

9x  = -45

x = -45/9

x  = -5.

EVALUATION

Solve the following :

1.  2 (x + 5) = 18                    2. 6 (2s – 7)  = 5s             3.  3x + 1 = 2(3x+5)

4.  8 (2d – 3)  = 3 (4d – 7)       5. (y + 8 ) + 2 (y + 1)  = 0.

New General Mathematics   chapt. 13 Ex 13d nos 1-20.

WEEKEND ASSIGNMENT

1.  Solve 3x + 9 = 117

(a) 38               (b) 36    (c) -36  (d) -38

2.   If -2r = 18 what is 4?

(a) -9               (b) 20              (c) 9   (d) -20

3.  solve 2 (x + 5)  = 16

(a) 13               (b) 10              (c)  8                (d) 3

4.  Solve x   = 5    (a) -15                    (b) 15    (c) 10             (d) -10

3

5.  If  x/5= ½  What is x?      (a) 2 ½                        (b) 2 2/3    (c) -2 ½       (d ) 2.
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