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 599: 51


$a=\left\{ -\dfrac{3}{5},5 \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ |3a-1|=|2a+4| ,$ use the definition of an absolute value equality. Then use the properties of equality 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} 3a-1=2a+4 \\\\\text{OR}\\\\ 3a-1=-(2a+4) .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} 3a-1=2a+4 \\\\ 3a-2a=4+1 \\\\ a=5 \\\\\text{OR}\\\\ 3a-1=-(2a+4) \\\\ 3a-1=-2a-4 \\\\ 3a+2a=-4+1 \\\\ 5a=-3 \\\\ a=-\dfrac{3}{5} .\end{array} Hence, $ a=\left\{ -\dfrac{3}{5},5 \right\} .$
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