College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 1 - Section 1.8 - Absolute Value Equations and Inequalities - 1.8 Exercises - Page 154: 44


$x=\left\{ \dfrac{7}{3},3 \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ |8-3x|-3=-2 ,$ use the properties of equality to isolate the absolute value expression. Then use the properties of absolute value equality. $\bf{\text{Solution Details:}}$ Using the properties of equality, the equation above is equivalent to \begin{array}{l}\require{cancel} |8-3x|=-2+3 \\\\ |8-3x|=1 .\end{array} Since for any $c\gt0$, $|x|=c$ implies $x=c \text{ or } x=-c,$ the equation above is equivalent to \begin{array}{l}\require{cancel} 8-3x=1 \\\\\text{OR}\\\\ 8-3x=-1 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} 8-3x=1 \\\\ -3x=1-8 \\\\ -3x=-7 \\\\ x=\dfrac{-7}{-3} \\\\ x=\dfrac{7}{3} \\\\\text{OR}\\\\ 8-3x=-1 \\\\ -3x=-1-8 \\\\ -3x=-9 \\\\ x=\dfrac{-9}{-3} \\\\ x=3 .\end{array} Hence, $ x=\left\{ \dfrac{7}{3},3 \right\} .$
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