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: 10


$h=\left\{ \dfrac{5}{2},\dfrac{7}{2} \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ 8|h-3|=4 ,$ 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} 8|h-3|=4 \\\\ |h-3|=\dfrac{4}{8} \\\\ |h-3|=\dfrac{1}{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} h-3=\dfrac{1}{2} \\\\\text{OR}\\\\ h-3=-\dfrac{1}{2} .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} h-3=\dfrac{1}{2} \\\\ h=\dfrac{1}{2}+3 \\\\ h=\dfrac{1}{2}+\dfrac{6}{2} \\\\ h=\dfrac{7}{2} \\\\\text{OR}\\\\ h-3=-\dfrac{1}{2} \\\\ h=-\dfrac{1}{2}+3 \\\\ h=-\dfrac{1}{2}+\dfrac{6}{2} \\\\ h=\dfrac{5}{2} .\end{array} If $h=\dfrac{7}{2},$ then \begin{array}{l}\require{cancel} 8|h-3|=4? \\\\ 8\left| \dfrac{7}{2}-3 \right|=4? \\\\ 8\left| \dfrac{7}{2}-\dfrac{6}{2} \right|=4? \\\\ 8\left| \dfrac{1}{2} \right|=4? \\\\ 8\left( \dfrac{1}{2} \right)=4? \\\\ 4=4 \text{ (TRUE)} .\end{array} If $h=\dfrac{5}{2},$ then \begin{array}{l}\require{cancel} 8|h-3|=4? \\\\ 8\left| \dfrac{5}{2}-3 \right|=4? \\\\ 8\left| \dfrac{5}{2}-\dfrac{6}{2} \right|=4? \\\\ 8\left| -\dfrac{1}{2} \right|=4? \\\\ 8\left( \dfrac{1}{2} \right)=4? \\\\ 4=4 \text{ (TRUE)} .\end{array} Hence, $ h=\left\{ \dfrac{5}{2},\dfrac{7}{2} \right\} .$
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