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: 24

Answer

$$z=-2$$

Work Step by Step

$$\frac{3}{4}(24 - 8z) - 16 = -\frac{2}{3}(6z - 9)$$ Simplify the left side of the equation by distributing $\frac{3}{4}$ to the terms inside the parentheses: $$\frac{3}{4}(24 - 8z) - 16$$ $$=[24(\frac{3}{4}) - 8z(\frac{3}{4})] - 16$$ $$=18-6z - 16$$ $$= 2-6z$$ Simplify the right side of the equation by distributing $-\frac{2}{3}$ to the terms inside the parentheses: $$-\frac{2}{3}(6z - 9)$$ $$=[6z(-\frac{2}{3})-9(-\frac{2}{3})]$$ $$= -4z + 6$$ Rewrite the equation using the simplified forms: $$2-6z = -4z + 6$$ Subtract $2$ from both sides: $$2-2-6z = -4z + 6 - 2$$ $$-6z= -4x + 4$$ Simplify: $$-2z = 4$$ Divide both sides by $-2$: $$z = -2$$ Check: $$\frac{3}{4}(24 - 8(-2)) - 16 = -\frac{2}{3}(6(-2) - 9)$$ $$\frac{3}{4}(24 + 16) - 16 = -\frac{2}{3}(-12 - 9)$$ $$\frac{3}{4}(40) - 16 = -\frac{2}{3}(-21)$$ $$30 - 16 = -(-14)$$ $$14 = 14 --> TRUE$$
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