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 - 9.3 Absolute-Value Equations and Inequalities - 9.3 Exercise Set - Page 600: 85


$-\dfrac{1}{2}\le x \le \dfrac{7}{2}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To find all $x$ such that $f(x) \le4 ,$ with $f(x)= |2x-3| ,$ solve the inequality \begin{array}{l}\require{cancel} |2x-3|\le4 .\end{array} Then graph the solution set. In the graph, a hollowed dot is used for $\lt$ or $\gt.$ A solid dot is used for $\le$ or $\ge.$ $\bf{\text{Solution Details:}}$ 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} -4\le 2x-3 \le4 .\end{array} Using the properties of inequality, the inequality above is equivalent to \begin{array}{l}\require{cancel} -4\le 2x-3 \le4 \\\\ -4+3\le 2x-3+3 \le4+3 \\\\ -1\le 2x \le 7 \\\\ -\dfrac{1}{2}\le \dfrac{2x}{{2}} \le \dfrac{7}{{2}} \\\\ -\dfrac{1}{2}\le x \le \dfrac{7}{2} .\end{array} Hence, the solution set $ -\dfrac{1}{2}\le x \le \dfrac{7}{2} .$
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