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: 35


$x=\left\{ -\dfrac{11}{2},\dfrac{13}{2} \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ \left| \dfrac{2x-1}{3} \right|=4 ,$ use the definition of an absolute value equality. Then use the properties of equality to isolate the variable. $\bf{\text{Solution Details:}}$ Since for any $c\gt0$, $|x|=c$ implies $x=c \text{ or } x=-c,$ the given equation is equivalent to \begin{array}{l}\require{cancel} \dfrac{2x-1}{3}=4 \\\\\text{OR}\\\\ \dfrac{2x-1}{3}=-4 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} \dfrac{2x-1}{3}=4 \\\\ 3\cdot\dfrac{2x-1}{3}=3\cdot4 \\\\ 2x-1=12 \\\\ 2x=12+1 \\\\ 2x=13 \\\\ x=\dfrac{13}{2} \\\\\text{OR}\\\\ \dfrac{2x-1}{3}=-4 \\\\ 3\cdot\dfrac{2x-1}{3}=3\cdot(-4) \\\\ 2x-1=-12 \\\\ 2x=-12+1 \\\\ 2x=-11 \\\\ x=-\dfrac{11}{2} .\end{array} Hence, $ x=\left\{ -\dfrac{11}{2},\dfrac{13}{2} \right\} .$
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