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: 36


$x=\left\{ -\dfrac{14}{5},\dfrac{22}{5}, \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ \left| \dfrac{4-5x}{6} \right|=3 ,$ 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{4-5x}{6}=3 \\\\\text{OR}\\\\ \dfrac{4-5x}{6}=-3 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} \dfrac{4-5x}{6}=3 \\\\ 6\cdot\dfrac{4-5x}{6}=6\cdot3 \\\\ 4-5x=18 \\\\ -5x=18-4 \\\\ -5x=14 \\\\ x=\dfrac{14}{-5} \\\\ x=-\dfrac{14}{5} \\\\\text{OR}\\\\ \dfrac{4-5x}{6}=-3 \\\\ 6\cdot\dfrac{4-5x}{6}=6\cdot(-3) \\\\ 4-5x=-18 \\\\ -5x=-18-4 \\\\ -5x=-22 \\\\ x=\dfrac{-22}{-5} \\\\ x=\dfrac{22}{5} .\end{array} Hence, $ x=\left\{ -\dfrac{14}{5},\dfrac{22}{5}, \right\} .$
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