College Algebra (11th Edition)

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

Chapter 1 - Review Exercises - Page 166: 114


$x=\left\{ -\dfrac{15}{13},-\dfrac{13}{29} \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ \left| \dfrac{8x-1}{3x+2} \right|=7 ,$ use the definition of absolute value equality. $\bf{\text{Solution Details:}}$ 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} \dfrac{8x-1}{3x+2}=7 \\\\\text{OR}\\\\ \dfrac{8x-1}{3x+2}=-7 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} \dfrac{8x-1}{3x+2}=7 \\\\ (3x+2)\left( \dfrac{8x-1}{3x+2}\right)=7(3x+2) \\\\ 1(8x-1)=7(3x+2) \\\\ 8x-1=21x+14 \\\\ 8x-21x=14+1 \\\\ -13x=15 \\\\ x=\dfrac{15}{-13} \\\\ x=-\dfrac{15}{13} \\\\\text{OR}\\\\ \dfrac{8x-1}{3x+2}=-7 \\\\ (3x+2)\left( \dfrac{8x-1}{3x+2}\right)=-7(3x+2) \\\\ 1(8x-1)=-7(3x+2) \\\\ 8x-1=-21x-14 \\\\ 8x+21x=-14+1 \\\\ 29x=-13 \\\\ x=-\dfrac{13}{29} .\end{array} Hence, the solutions are $ x=\left\{ -\dfrac{15}{13},-\dfrac{13}{29} \right\} .$
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