## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$x\le2 \text{ or } x\ge8$ $\bf{\text{Solution Outline:}}$ To solve the given inqeuality, $12-|x-5|\le9 ,$ use the properties of inequality to isolate the absolute value expression. Then use the definition of a greater than absolute value inequality and solve each resulting inequality. Finally, graph the solution set. In the graph, a hollowed dot is used for $\lt$ or $\gt.$ A solid dot is used for $\le$ or $\ge.$ $\bf{\text{Solution Details:}}$ Using the properties of inequality to isolate the absolute value expression, the given is equivalent to \begin{array}{l}\require{cancel} 12-|x-5|\le9 \\\\ -|x-5|\le9-12 \\\\ -|x-5|\le-3 .\end{array} Dividing both sides by a negative number (and consequently reversing the inequality symbol) results to \begin{array}{l}\require{cancel} -|x-5|\le-3 \\\\ \dfrac{-|x-5|}{-1}\le\dfrac{-3}{-1} \\\\ |x-5|\ge3 .\end{array} 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} x-5\ge3 \\\\\text{OR}\\\\ x-5\le-3 .\end{array} Solving each inequality results to \begin{array}{l}\require{cancel} x-5\ge3 \\\\ x\ge3+5 \\\\ x\ge8 \\\\\text{OR}\\\\ x-5\le-3 \\\\ x\le-3+5 \\\\ x\le2 .\end{array} Hence, the solution set is $x\le2 \text{ or } x\ge8 .$