Intermediate Algebra (12th Edition)

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

Chapter 4 - Chapters R-4 - Cumulative Review Exercises: 9


$x=\left\{ -\dfrac{1}{3},1 \right\}$

Work Step by Step

Since for any $c\gt0$, $|x|=c$ implies $x=c$ or $x=-c$, the given equation, $ |3x-1|=2 ,$ is equivalent to \begin{array}{l}\require{cancel} 3x-1=2 \text{ OR } 3x-1=-2 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} 3x-1=2 \\\\ 3x=2+1 \\\\ 3x=3 \\\\ x=\dfrac{3}{3} \\\\ x=1 \\\\\text{ OR }\\\\ 3x-1=-2 \\\\ 3x=-2+1 \\\\ 3x=-1 \\\\ x=-\dfrac{1}{3} .\end{array} Hence, the solutions are $ x=\left\{ -\dfrac{1}{3},1 \right\} .$
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