Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 2 - Cumulative Review - Page 115: 44

Answer

$y=-36$ and $y=24$

Work Step by Step

We are given that $|\frac{y}{3}+2|=10$. So, $\frac{y}{3}+2=10$ and $\frac{y}{3}+2=-10$. We can solve both of these equations for y. For $\frac{y}{3}+2=10$, subtract 2 from both sides of the equation. $\frac{y}{3}=8$ Multiply both sides of the equation by 3. y=24 For $\frac{y}{3}+2=-10$, subtract 2 from both sides of the equation. $\frac{y}{3}=-12$ Multiply both sides of the equation by 3. y=-36
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