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

$-16\le x \le 8$
$\bf{\text{Solution Outline:}}$ To solve the given inequality, $\left| \dfrac{x+4}{6} \right|\le2 ,$ use the definition of a less than (less than or equal to) absolute value inequality. Then use the properties of inequality to isolate the variable. $\bf{\text{Solution Details:}}$ 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} -2\le \dfrac{x+4}{6} \le2 .\end{array} Using the properties of inequality, the inequality above is equivalent to \begin{array}{l}\require{cancel} -2\le \dfrac{x+4}{6} \le2 \\\\ 6\cdot(-2)\le 6\cdot\dfrac{x+4}{6} \le6\cdot2 \\\\ -12\le x+4 \le 12 \\\\ -12-4\le x+4-4 \le 12-4 \\\\ -16\le x \le 8 .\end{array} Hence, the solution set $-16\le x \le 8 .$