Intermediate Algebra: Connecting Concepts through Application

$a=\left\{ -8,24 \right\}$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $3|a-8|-4=44 ,$ 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|a-8|=44+4 \\\\ 3|a-8|=48 \\\\ |a-8|=\dfrac{48}{3} \\\\ |a-8|=16 .\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} a-8=16 \\\\\text{OR}\\\\ a-8=-16 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} a-8=16 \\\\ a=16+8 \\\\ a=24 \\\\\text{OR}\\\\ a-8=-16 \\\\ a=-16+8 \\\\ a=-8 .\end{array} If $a=24,$ then \begin{array}{l}\require{cancel} 3|a-8|-4=44? \\\\ 3|24-8|-4=44? \\\\ 3|16|-4=44? \\\\ 3(16)-4=44? \\\\ 48-4=44 \\\\ 44=44 \text{ (TRUE)} .\end{array} If $a=-8,$ then \begin{array}{l}\require{cancel} 3|a-8|-4=44? \\\\ 3|-8-8|-4=44? \\\\ 3|-16|-4=44? \\\\ 3(16)-4=44? \\\\ 48-4=44? \\\\ 44=44 \text{ (TRUE)} .\end{array} Hence, $a=\left\{ -8,24 \right\} .$