Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 2 - Systems of Linear Equations and Inequalities - 2.5 Absolute Value Equations and Inequalities - 2.5 Exercises - Page 186: 15


$x=\left\{ \dfrac{31}{3},\dfrac{41}{3} \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ -3|x-12|=-5 ,$ isolate first the absolute value expression. Then use the definition of absolute value equality. Do checking of the solution/s. $\bf{\text{Solution Details:}}$ Using the properties of equality, the given equation is equivalent to \begin{array}{l}\require{cancel} -3|x-12|=-5 \\\\ |x-12|=\dfrac{-5}{-3} \\\\ |x-12|=\dfrac{5}{3} .\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} x-12=\dfrac{5}{3} \\\\\text{OR}\\\\ x-12=-\dfrac{5}{3} .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} x-12=\dfrac{5}{3} \\\\ x=\dfrac{5}{3}+12 \\\\ x=\dfrac{5}{3}+\dfrac{36}{3} \\\\ x=\dfrac{41}{3} \\\\\text{OR}\\\\ x-12=-\dfrac{5}{3} \\\\ x=-\dfrac{5}{3}+12 \\\\ x=-\dfrac{5}{3}+\dfrac{36}{3} \\\\ x=\dfrac{31}{3} .\end{array} If $x=\dfrac{41}{3},$ then \begin{array}{l}\require{cancel} -3|x-12|=-5? \\\\ -3\left| \dfrac{41}{3}-12 \right|=-5? \\\\ -3\left| \dfrac{41}{3}-\dfrac{36}{3} \right|=-5? \\\\ -3\left| \dfrac{5}{3} \right|=-5? \\\\ -3\left( \dfrac{5}{3} \right)=-5? \\\\ -5=-5 \text{ (TRUE)} .\end{array} If $x=\dfrac{31}{3},$ then \begin{array}{l}\require{cancel} -3|x-12|=-5? \\\\ -3\left| \dfrac{31}{3}-12 \right|=-5? \\\\ -3\left| \dfrac{31}{3}-\dfrac{36}{3} \right|=-5? \\\\ -3\left| -\dfrac{5}{3} \right|=-5? \\\\ -3\left( \dfrac{5}{3} \right)=-5? \\\\ -5=-5 \text{ (TRUE)} .\end{array} Hence, $ x=\left\{ \dfrac{31}{3},\dfrac{41}{3} \right\} .$
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