## Intermediate Algebra: Connecting Concepts through Application

$x=\left\{ -14,10 \right\}$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $|x+2|+6=18 ,$ 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} |x+2|+6=18 \\\\ |x+2|=18-6 \\\\ |x+2|=12 .\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} x+2=12 \\\\\text{OR}\\\\ x+2=-12 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} x+2=12 \\\\ x=12-2 \\\\ x=10 \\\\\text{OR}\\\\ x+2=-12 \\\\ x=-12-2 \\\\ x=-14 .\end{array} If $x=10,$ then \begin{array}{l}\require{cancel} |x+2|+6=18? \\\\ |10+2|+6=18? \\\\ |12|+6=18? \\\\ 12+6=18? \\\\ 18=18 \text{ (TRUE)} .\end{array} If $x=-14,$ then \begin{array}{l}\require{cancel} |x+2|+6=18? \\\\ |-14+2|+6=18? \\\\ |-12|+6=18? \\\\ 12+6=18? \\\\ 18=18 \text{ (TRUE)} .\end{array} Hence, $x=\left\{ -14,10 \right\} .$