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

$x\le-8 \text{ or } x\ge0$ $\bf{\text{Solution Outline:}}$ To solve the given inqeuality, $9-|x+4|\le5 ,$ 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} 9-|x+4|\le5 \\\\ -|x+4|\le5-9 \\\\ -|x+4|\le-4 .\end{array} Dividing both sides by a negative number (and consequently reversing the inequality symbol) results to \begin{array}{l}\require{cancel} -|x+4|\le-4 \\\\ \dfrac{-|x+4|}{-1}\le\dfrac{-4}{-1} \\\\ |x+4|\ge4 .\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+4\ge4 \\\\\text{OR}\\\\ x+4\le-4 .\end{array} Solving each inequality results to \begin{array}{l}\require{cancel} x+4\ge4 \\\\ x\ge4-4 \\\\ x\ge0 \\\\\text{OR}\\\\ x+4\le-4 \\\\ x\le-4-4 \\\\ x\le-8 .\end{array} Hence, the solution set is $x\le-8 \text{ or } x\ge0 .$