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: 86

Answer

$-1\le x \le \dfrac{1}{5}$
1515750315

Work Step by Step

$\bf{\text{Solution Outline:}}$ To find all $x$ such that $f(x) \le3 ,$ with $f(x)= |5x+2| ,$ solve the inequality \begin{array}{l}\require{cancel} |5x+2|\le3 .\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} -3\le 5x+2 \le3 .\end{array} Using the properties of inequality, the inequality above is equivalent to \begin{array}{l}\require{cancel} -3\le 5x+2 \le3 \\\\ -3-2\le 5x+2-2 \le3-2 \\\\ -5\le 5x \le 1 \\\\ -\dfrac{5}{5}\le \dfrac{5x}{5} \le \dfrac{1}{5} \\\\ -1\le x \le \dfrac{1}{5} .\end{array} Hence, the solution set $ -1\le x \le \dfrac{1}{5} .$
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