## Intermediate Algebra (12th Edition)

$\left( -\infty, -2 \right) \cup \left( 8,\infty \right)$
$\bf{\text{Solution Outline:}}$ To solve the given inequality, $|3-x|\gt5 ,$ use the definition of absolute value inequalities. Then use the properties of inequality to isolate the variable. For the interval notation, use a parenthesis for the symbols $\lt$ or $\gt.$ Use a bracket for the symbols $\le$ or $\ge.$ For graphing inequalities, use a hollowed dot for the symbols $\lt$ or $\gt.$ Use a solid dot for the symbols $\le$ or $\ge.$ $\bf{\text{Solution Details:}}$ Since for any $c\gt0$, $|x|\gt c$ implies $x\gt c \text{ or } x\lt-c$ (which is equivalent to $|x|\ge c$ implies $x\ge c \text{ or } x\le-c$), the inequality above is equivalent to \begin{array}{l}\require{cancel} 3-x\gt5 \\\\\text{OR}\\\\ 3-x\lt-5 .\end{array} Using the properties of inequality to isolate the variable results to \begin{array}{l}\require{cancel} -x\gt5-3 \\\\ -x\gt2 \\\\\text{OR}\\\\ 3-x\lt-5 \\\\ -x\lt-5-3 \\\\ -x\lt-8 .\end{array} Dividing by a negative number (and consequently reversing the sign), the inequality above is equivalent to \begin{array}{l}\require{cancel} -x\gt2 \\\\ x\lt\dfrac{2}{-1} \\\\ x\lt-2 \\\\\text{OR}\\\\ -x\lt-8 \\\\ x\gt\dfrac{-8}{-1} \\\\ x\gt8 .\end{array} In interval notation, the solution set is $\left( -\infty, -2 \right) \cup \left( 8,\infty \right) .$ The colored graph is the graph of the solution set.