Intermediate Algebra (12th Edition)

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

Chapter 5 - Chapters R-5 - Cumulative Review Exercises - Page 363: 10


$x=\left\{ -\dfrac{16}{5},2 \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given inequality, $ |5x+3|-10=3 ,$ use the properties of equality to isolate the absolute value expression. Then use the definition of absolute value to solve for the value/s of the variable. $\bf{\text{Solution Details:}}$ Using the properties of equality, the equation above is equivalent to \begin{array}{l}\require{cancel} |5x+3|=3+10 \\\\ |5x+3|=13 .\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} 5x+3=13 \\\\\text{OR}\\\\ 5x+3=-13 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} 5x+3=13 \\\\ 5x=13-3 \\\\ 5x=10 \\\\ x=\dfrac{10}{5} \\\\ x=2 \\\\\text{OR}\\\\ 5x+3=-13 \\\\ 5x=-13-3 \\\\ 5x=-16 \\\\ x=-\dfrac{16}{5} .\end{array} Hence, $ x=\left\{ -\dfrac{16}{5},2 \right\} .$
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