## Intermediate Algebra (12th Edition)

$\left[ 3,13 \right]$
$\bf{\text{Solution Outline:}}$ To solve the given inequality, $|-8+x| \le 5 ,$ use the definition of absolute value inequalities. Use the properties of inequalities 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 $c\gt0$, $|x|\lt c$ implies $-c\lt x\lt c$ (or $|x|\le c$ implies $-c\le x\le c$), the inequality above is equivalent to \begin{array}{l}\require{cancel} -5 \le -8+x \le 5 .\end{array} Using the properties of inequality, the inequality above is equivalent to \begin{array}{l}\require{cancel} -5+8 \le -8+x+8 \le 5+8 \\\\ 3 \le x \le 13 .\end{array} In interval notation, the solution set is $\left[ 3,13 \right] .$ The colored graph is the graph of the solution set.