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: 23

Answer

$x=\left\{ -10,-2 \right\}$

Work Step by Step

Since for any $a\gt0$, $|x|=a$ implies $x=a$ OR $x=-a$, then the given equation, $ \left|\dfrac{1}{2}x+3\right|=2 ,$ is equivalent to \begin{array}{l}\require{cancel} \dfrac{1}{2}x+3=2 \text{ OR } \dfrac{1}{2}x+3=-2 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} \dfrac{1}{2}x+3=2 \\\\ 2\left(\dfrac{1}{2}x+3\right)=(2)2 \\\\ x+6=4 \\\\ x=4-6 \\\\ x=-2 \\\\\text{ OR }\\\\ \dfrac{1}{2}x+3=-2 \\\\ 2\left(\dfrac{1}{2}x+3\right)=(-2)2 \\\\ x+6=-4 \\\\ x=-4-6 \\\\ x=-10 .\end{array} Hence, the solutions are $ x=\left\{ -10,-2 \right\} .$
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