Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 9 - Inequalities and Problem Solving - Review Exercises: Chapter 9 - Page 623: 37


$-16\le x \le 8$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given inequality, $ \left| \dfrac{x+4}{6} \right|\le2 ,$ use the definition of a less than (less than or equal to) absolute value inequality. Then use the properties of inequality to isolate the variable. $\bf{\text{Solution Details:}}$ Since for any $c\gt0$, $|x|\lt c$ implies $-c\lt x\lt c$ (or $|x|\le c$ implies $-c\le x\le c$), the inequality above is equivalent to \begin{array}{l}\require{cancel} -2\le \dfrac{x+4}{6} \le2 .\end{array} Using the properties of inequality, the inequality above is equivalent to \begin{array}{l}\require{cancel} -2\le \dfrac{x+4}{6} \le2 \\\\ 6\cdot(-2)\le 6\cdot\dfrac{x+4}{6} \le6\cdot2 \\\\ -12\le x+4 \le 12 \\\\ -12-4\le x+4-4 \le 12-4 \\\\ -16\le x \le 8 .\end{array} Hence, the solution set $ -16\le x \le 8 .$
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