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: 16


$y=\left\{ -\dfrac{106}{5}, -\dfrac{34}{5} \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given equation, $ -5|y+14|=-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} -5|y+14|=-36 \\\\ \dfrac{-5|y+14|}{-5}=\dfrac{-36}{-5} \\\\ |y+14|=\dfrac{36}{5} .\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} y+14=\dfrac{36}{5} \\\\\text{OR}\\\\ y+14=-\dfrac{36}{5} .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} y+14=\dfrac{36}{5} \\\\ y=\dfrac{36}{5}-14 \\\\ y=\dfrac{36}{5}-\dfrac{70}{5} \\\\ y=-\dfrac{34}{5} \\\\\text{OR}\\\\ y+14=-\dfrac{36}{5} \\\\ y=-\dfrac{36}{5}-14 \\\\ y=-\dfrac{36}{5}-\dfrac{70}{5} \\\\ y=-\dfrac{106}{5} .\end{array} If $y=-\dfrac{34}{5},$ then \begin{array}{l}\require{cancel} -5|y+14|=-36? \\\\ -5\left| -\dfrac{34}{5}+14 \right|=-36? \\\\ -5\left| -\dfrac{34}{5}+\dfrac{70}{5} \right|=-36? \\\\ -5\left| \dfrac{36}{5} \right|=-36? \\\\ -5\left( \dfrac{36}{5} \right)=-36? \\\\ -36=-36 \text{ (TRUE)} .\end{array} If $y=-\dfrac{106}{5},$ then \begin{array}{l}\require{cancel} -5|y+14|=-36? \\\\ -5\left| -\dfrac{106}{5}+14 \right|=-36? \\\\ -5\left| -\dfrac{106}{5}+\dfrac{70}{5} \right|=-36? \\\\ -5\left| -\dfrac{36}{5} \right|=-36? \\\\ -5\left( \dfrac{36}{5} \right)=-36? \\\\ -36=-36 \text{ (TRUE)} .\end{array} Hence, $ y=\left\{ -\dfrac{106}{5}, -\dfrac{34}{5} \right\} .$
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