Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 1 - Section 1.7 - Absolute Value Equations and Inequalities - 1.7 Exercises - Page 119: 69


$x=\left\{ -\dfrac{5}{3},\dfrac{11}{3} \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ |3(x-1)|=8 ,$ use the Distributive Property and the definition of absolute value equality. Then use the properties of equality to isolate the variable. $\bf{\text{Solution Details:}}$ Using the Distributive Property, the given equation is equivalent to \begin{array}{l}\require{cancel} |3(x)+3(-1)|=8 \\\\ |3x-3|=8 .\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} 3x-3=8 \\\\\text{OR}\\\\ 3x-3=-8 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} 3x-3=8 \\\\ 3x=8+3 \\\\ 3x=11 \\\\ x=\dfrac{11}{3} \\\\\text{OR}\\\\ 3x-3=-8 \\\\ 3x=-8+3 \\\\ 3x=-5 \\\\ x=-\dfrac{5}{3} .\end{array} Hence, $ x=\left\{ -\dfrac{5}{3},\dfrac{11}{3} \right\} .$
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