Page 610 - Basic College Mathematics with Early Integers

S E C T I O N 8.4 I SOLVING EQUATIONS USING ADDITION AND MULTIPLICATION PROPERTIES 587

Example 2 Solve: 17 - x + 3 = 15 - (-6) PRACTICE 2
Solve:
Solution: First we simplify each side of the equation. 1 - y + 9 = 20 - 1-252

17 - x + 3 = 15 - (-6)
20 - x = 21 Combine like terms on each side of the equation.
Next, we get the variable term alone on one side of the equation.
20 - x - 20 = 21 - 20 Subtract 20 from both sides.
-1x = 1 Simplify. Recall that -x means -1x.
-1x 1
= Divide both sides by -1.
-1 -1
#
1 x =-1 Simplify.
x =-1
Check:

20 - x = 21

20 - 1-12 21
21 21 True
The solution is -1.
Work Practice 2

2
Example 3 Solve: 1 = x + 7 PRACTICE 3
3
3
Solution: Subtract 7 from both sides to get the variable term alone. Solve: 11 = y + 20
4
2
1 - 7 = x + 7 - 7 Subtract 7 from both sides.
3
2
-6 = x Simplify.
3
3 3 2 3
# -6 = # x Multiply both sides by .
2 2 3 2
-3
3 -6
#
# = 1 x Simplify.
2 1 Don’t forget that
1
-9 = x Simplify. we can get the variable alone
on either side of the equation.
Check to see that the solution is -9.
Work Practice 3

If an equation contains variable terms on both sides, we use the addition prop-
erty of equality to get all the variable terms on one side and all the constants, or
numbers, on the other side.

Example 4 Solve: 3a - 6 = a + 4 PRACTICE 4
Solve: 9x - 12 = x + 4
Solution:
3a - 6 = a + 4
3a - 6 + 6 = a + 4 + 6 Add 6 to both sides.
3a = a + 10 Simplify.
3a - a = a + 10 - a Subtract a from both sides.
Answers
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