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: 9


$r=\{ -29,7 \}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ 2|r+11|=36 ,$ 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} 2|r+11|=36 \\\\ |r+11|=\dfrac{36}{2} \\\\ |r+11|=18 .\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} r+11=18 \\\\\text{OR}\\\\ r+11=-18 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} r+11=18 \\\\ r=18-11 \\\\ r=7 \\\\\text{OR}\\\\ r+11=-18 \\\\ r=-18-11 \\\\ r=-29 .\end{array} If $r=7,$ then \begin{array}{l}\require{cancel} 2|r+11|=36? \\\\ 2|7+11|=36? \\\\ 2|18|=36? \\\\ 2(18)=36? \\\\ 36=36 \text{ (TRUE)} .\end{array} If $r=-29,$ then \begin{array}{l}\require{cancel} 2|r+11|=36? \\\\ 2|-29+11|=36? \\\\ 2|-18|=36? \\\\ 2(18)=36? \\\\ 36=36 \text{ (TRUE)} .\end{array} Hence, $ r=\{ -29,7 \} .$
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