College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 1 - Section 1.6 - Equations and Inequalities Involving Absolute Value - 1.6 Assess Your Understanding - Page 133: 21


The solution set is {$-\frac{36}{5}, \frac{24}{5}$}

Work Step by Step

$|\frac{x}{3}+\frac{2}{5}|=2$ $\frac{x}{3}+\frac{2}{5}=2$ or $\frac{x}{3}+\frac{2}{5}=-2$ $\frac{x}{3}=2-\frac{2}{5}$ or $\frac{x}{3}=-2-\frac{2}{5}$ $\frac{x}{3}=\frac{10-2}{5}$ or $\frac{x}{3}=\frac{-10-2}{5}$ $\frac{x}{3}=\frac{8}{5}$ or $\frac{x}{3}=-\frac{12}{5}$ $x=\frac{8}{5}\cdot3$ or $x=-\frac{12}{5}\cdot3$ $x=\frac{24}{5}$ or $x=-\frac{36}{5}$ The solution set is {$-\frac{36}{5}, \frac{24}{5}$}
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