Page 611 - Basic College Mathematics with Early Integers

588 C HAPTE R 8 I INTRODUCTION TO ALGEBRA

2a = 10 Simplify.
2a 10
= Divide both sides by 2.
2 2
a = 5 Simplify.
Check to see that the solution is 5.
Work Practice 4

PRACTICE 5 Example 5 Solve: 7x + 3.2 = 4x - 1.6

Solve: 8x + 4.2 = 10x - 11.6
Solution:
7x + 3.2 = 4x - 1.6
7x + 3.2 - 3.2 = 4x - 1.6 - 3.2 Subtract 3.2 from both sides.
7x = 4x - 4.8 Simplify.
7x - 4x = 4x - 4.8 - 4x Subtract 4x from both sides.
3x =-4.8 Simplify.
3x -4.8
= Divide both sides by 3.
3 3
x =-1.6 Simplify.
Check to see that -1.6 is the solution.
Work Practice 5

Objective Solving Equations Containing Parentheses
If an equation contains parentheses, we must first use the distributive property to
remove them.

PRACTICE 6 Example 6 Solve: 71x - 22 = 9x - 6
Solve: 61a - 52 = 7a - 13
Solution: First we apply the distributive property.

7(x-2)=9x-6
7x - 14 = 9x - 6 Apply the distributive property.

Next, we move variable terms to one side of the equation and constants to the
other side.
7x - 14 - 9x = 9x - 6 - 9x Subtract 9x from both sides.

-2x - 14 =-6 Simplify.
-2x - 14 + 14 =-6 + 14 Add 14 to both sides.
-2x = 8 Simplify.
-2x 8
= Divide both sides by -2.
Copyright 2012 Pearson Education, Inc.
-2 -2
x =-4 Simplify.
Check to see that -4 is the solution.
Work Practice 6

Answers
5. 7.9 6. -17