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: 45


$x=\left\{ 2,4 \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ |6-2x|+1=3 ,$ 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} |6-2x|=3-1 \\\\ |6-2x|=2 .\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} 6-2x=2 \\\\\text{OR}\\\\ 6-2x=-2 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} 6-2x=2 \\\\ -2x=2-6 \\\\ -2x=-4 \\\\ x=\dfrac{-4}{-2} \\\\ x=2 \\\\\text{OR}\\\\ 6-2x=-2 \\\\ -2x=-2-6 \\\\ -2x=-8 \\\\ x=\dfrac{-8}{-2} \\\\ x=4 .\end{array} Hence, $ x=\left\{ 2,4 \right\} .$
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