Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 1 - Section 1.4 - Solving Linear Equations - Exercise Set - Page 50: 23

Answer

$$z = 5/2$$

Work Step by Step

$\frac{1}{2}(4z + 8) - 16 = -\frac{2}{3}(9z - 12)$ Expand the left side of the equation: $\frac{1}{2}(4z + 8) - 16$ = $(2z + 4) - 16$ = $2z - 12$ Expand the right side of the equation: $-\frac{2}{3}(9z - 12)$ = $-6z + 8$ Rewrite the equation: $2z - 12$ = $-6z + 8$ Add 12 to both sides: $$2z - 12 + 12 = -6z + 8 + 12$$ $$2z = -6z + 20$$ Simplify: $$8z = 20$$ Divide both sides by $8$: $$z = 5/2$$
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