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

Published by Pearson

# Chapter 9 - Inequalities and Problem Solving - 9.3 Absolute-Value Equations and Inequalities - 9.3 Exercise Set - Page 600: 89

#### Answer

$-4\lt x \lt 5$ #### Work Step by Step

$\bf{\text{Solution Outline:}}$ To find all $x$ such that $f(x) \lt16 ,$ with $f(x)= 7+|2x-1| ,$ solve the inequality \begin{array}{l}\require{cancel} 7+|2x-1|\lt16 .\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 $\ge.$ $\bf{\text{Solution Details:}}$ Isolating the absolute value expression results to \begin{array}{l}\require{cancel} 7+|2x-1|\lt16 \\\\ |2x-1|\lt16-7 \\\\ |2x-1|\lt9 .\end{array} 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} -9\lt 2x-1 \lt9 .\end{array} Using the properties of inequality, the inequality above is equivalent to \begin{array}{l}\require{cancel} -9\lt 2x-1 \lt9 \\\\ -9+1\lt 2x-1+1 \lt9+1 \\\\ -8\lt 2x \lt 10 \\\\ -\dfrac{8}{2}\lt \dfrac{2x}{2} \lt \dfrac{10}{2} \\\\ -4\lt x \lt 5 .\end{array} Hence, the solution set $-4\lt x \lt 5 .$

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