## College Algebra (11th Edition)

$\left[ -1,-\dfrac{1}{2} \right]$
Since for any $a\gt0$, $|x|\le a$ implies $-a\le x \le a$ then the solution to the given inequality, $\left| \dfrac{2}{3}x+\dfrac{1}{2} \right|\le \dfrac{1}{6} ,$ is \begin{array}{l}\require{cancel} -\dfrac{1}{6}\le \dfrac{2}{3}x+\dfrac{1}{2} \le \dfrac{1}{6} \\\\ 6\left( -\dfrac{1}{6} \right)\le 6\left( \dfrac{2}{3}x+\dfrac{1}{2} \right) \le 6\left( \dfrac{1}{6} \right) \\\\ -1\le 4x+3 \le 1 \\\\ -1-3\le 4x+3-3 \le 1-3 \\\\ -4\le 4x \le -2 \\\\ -\dfrac{4}{4}\le \dfrac{4x}{4} \le \dfrac{-2}{4} \\\\ -1\le x \le -\dfrac{1}{2} .\end{array} Hence, the solution set is the interval $\left[ -1,-\dfrac{1}{2} \right] .$