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: 20

Answer

$x=\left\{ 18,66 \right\}$

Work Step by Step

Since for any $a\gt0$, $|x|=a$ implies $x=a$ OR $x=-a$, then the given equation, $ \left|14-\dfrac{1}{3}x\right|=8 ,$ is equivalent to \begin{array}{l}\require{cancel} 14-\dfrac{1}{3}x=8 \text{ OR } 14-\dfrac{1}{3}x=-8 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} 14-\dfrac{1}{3}x=8 \\\\ -\dfrac{1}{3}x=8-14 \\\\ -\dfrac{1}{3}x=-6 \\\\ -3\left(-\dfrac{1}{3}x\right)=(-6)(-3) \\\\ x=18 \\\\\text{ OR }\\\\ 14-\dfrac{1}{3}x=-8 \\\\ -\dfrac{1}{3}x=-8-14 \\\\ -\dfrac{1}{3}x=-22 \\\\ -3\left(-\dfrac{1}{3}x\right)=(-22)(-3) \\\\ x=66 .\end{array} Hence, the solutions are $ x=\left\{ 18,66 \right\} .$
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