## Intermediate Algebra: Connecting Concepts through Application

$r=\left\{ -3,-\dfrac{1}{3} \right\}$
$\bf{\text{Solution Outline:}}$ To solve the given inequality, $-2|3r+5|+15=7 ,$ isolate first the absolute value expression. Then use the definition of absolute value to analyze the solution. $\bf{\text{Solution Details:}}$ Using the properties of inequality, the given is equivalent to \begin{array}{l}\require{cancel} -2|3r+5|+15=7 \\\\ -2|3r+5|=7-15 \\\\ -2|3r+5|=-8 \\\\ |3r+5|=\dfrac{-8}{-2} \\\\ |3r+5|=4 .\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} 3r+5=4 \\\\\text{OR}\\\\ 3r+5=-4 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} 3r+5=4 \\\\ 3r=4-5 \\\\ 3r=-1 \\\\ r=-\dfrac{1}{3} \\\\\text{OR}\\\\ 3r+5=-4 \\\\ 3r=-4-5 \\\\ 3r=-9 \\\\ r=-\dfrac{9}{3} \\\\ r=-3 .\end{array} If $r=-\dfrac{1}{3},$ then \begin{array}{l}\require{cancel} -2|3r+5|+15=7? \\\\ -2\left| 3\left( -\dfrac{1}{3} \right)+5 \right|+15=7? \\\\ -2\left| -1+5 \right|+15=7? \\\\ -2\left| 4 \right|+15=7? \\\\ -2(4)+15=7? \\\\ -8+15=7? \\\\ 7=7 \text{ (TRUE)} .\end{array} If $r=-3,$ then \begin{array}{l}\require{cancel} -2|3r+5|+15=7? \\\\ -2\left| 3\left( -3 \right)+5 \right|+15=7? \\\\ -2\left| -9+5 \right|+15=7? \\\\ -2\left| -4 \right|+15=7? \\\\ -2(4)+15=7? \\\\ -8+15=7? \\\\ 7=7 \text{ (TRUE)} .\end{array} Hence, $r=\left\{ -3,-\dfrac{1}{3} \right\} .$