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: 22


The solution set is {$-\frac{4}{3},\frac{8}{3}$}

Work Step by Step

$|\frac{x}{2}-\frac{1}{3}|=1$ $\frac{x}{2}-\frac{1}{3}=1$ or $\frac{x}{2}-\frac{1} {3}=-1$ $\frac{x}{2}=1+\frac{1}{3}$ or $\frac{x}{2}=-1+\frac{1} {3}$ $\frac{x}{2}=\frac{3+1}{3}$ or $\frac{x}{2}=\frac{-3+1}{3}$ $\frac{x}{2}=\frac{4}{3}$ or $\frac{x}{2}=-\frac{2}{3}$ $x=\frac{4}{3}\cdot2$ or $x=-\frac{2}{3}\cdot2$ $x=\frac{8}{3}$ or $x=-\frac{4}{3}$ The solution set is {$-\frac{4}{3},\frac{8}{3}$}
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