Page 601 - Basic College Mathematics with Early Integers

578 C HAPTE R 8 I INTRODUCTION TO ALGEBRA

Check: To check, we replace x with -3 in the original equation.
-5x = 15 Original equation

-51-32 15 Let x =-3.
15 15 True
The solution is -3.
Work Practice 1

PRACTICE 2 Example 2 Solve: -8 = 2y
Solve: -16 = 8x
Solution: To get y alone, we divide both sides of the equation by 2.
-8 = 2y

-8 2y
= Divide both sides by 2.
2 2
#
-4 = 1 y or y =-4
Check to see that -4 is the solution.

Work Practice 2

PRACTICE 3 Example 3 Solve: -1.2x =-36
Solve: -0.3y =-27
Solution: We divide both sides of the equation by the coefficient of x, which is
-1.2.
-1.2x =-36
-1.2x -36
=
-1.2 -1.2
#
1 x = 30
x = 30
Check to see that 30 is the solution.

Work Practice 3

3
PRACTICE 4 Example 4 Solve: a = 9
5
5
Solve: b = 25
7 Solution: Recall that the product of a number and its reciprocal is 1. To get a
5 3
alone, then, we multiply both sides by , the reciprocal of .
3 5
3
a = 9
5
5 3 5 5
# a = # 9 Multiply both sides by .
3 5 3 3
3
5 # 9 Copyright 2012 Pearson Education, Inc.
#
1 a = Multiply.
#
3 1
1
Answers a = 15 Simplify.
2. -2 3. 90 4. 35