Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 1 - Section 1.7 - Absolute Value Equations and Inequalities - 1.7 Exercises: 24

Answer

$x=\left\{ -6,9 \right\}$

Work Step by Step

Since for any $a\gt0$, $|x|=a$ implies $x=a$ OR $x=-a$, then the given equation, $ \left|\dfrac{2}{3}x-1\right|=5 ,$ is equivalent to \begin{array}{l}\require{cancel} \dfrac{2}{3}x-1=5 \text{ OR } \dfrac{2}{3}x-1=-5 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} \dfrac{2}{3}x-1=5 \\\\ 3\left(\dfrac{2}{3}x-1\right)=(5)3 \\\\ 2x-3=15 \\\\ 2x=15+3 \\\\ 2x=18 \\\\ x=\dfrac{18}{2} \\\\ x=9 \\\\\text{ OR }\\\\ 3\left(\dfrac{2}{3}x-1\right)=(-5)3 \\\\ 2x-3=-15 \\\\ 2x=-15+3 \\\\ 2x=-12 \\\\ x=-\dfrac{12}{2} \\\\ x=-6 .\end{array} Hence, the solutions are $ x=\left\{ -6,9 \right\} .$
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