## 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 599: 49

#### Answer

$x=-\dfrac{1}{2}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $|x+4|=|x-3| ,$ use the definition of absolute equality. Then use the properties of anequality to isolate the variable. $\bf{\text{Solution Details:}}$ Since $|x|=|y|$ implies $x=y \text{ or } x=-y,$ the equation above is equivalent to \begin{array}{l}\require{cancel} x+4=x-3 \\\\\text{OR}\\\\ x+4=-(x-3) .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} x+4=x-3 \\\\ x-x=-3-4 \\\\ 0=-7 \text{ (FALSE)} \\\\\text{OR}\\\\ x+4=-(x-3) \\\\ x+4=-x+3 \\\\ x+x=3-4 \\\\ 2x=-1 \\\\ x=-\dfrac{1}{2} .\end{array} Hence, $x=-\dfrac{1}{2} .$

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