#### Answer

The solution is $\Big[-1,-\dfrac{1}{2}\Big]$

#### Work Step by Step

$\Big|\dfrac{2}{3}x+\dfrac{1}{2}\Big|\le\dfrac{1}{6}$
Solving this absolute value inequality is equivalent to solving the following inequality:
$-\dfrac{1}{6}\le\dfrac{2}{3}x+\dfrac{1}{2}\le\dfrac{1}{6}$
$\textbf{Solve the inequality shown above:}$
$-\dfrac{1}{6}\le\dfrac{2}{3}x+\dfrac{1}{2}\le\dfrac{1}{6}$
Multiply the whole inequality by $6$:
$6\Big(-\dfrac{1}{6}\le\dfrac{2}{3}x+\dfrac{1}{2}\le\dfrac{1}{6}\Big)$
$-1\le4x+3\le1$
Subtract $3$ from all three parts of the inequality:
$-1-3\le4x+3-3\le1-3$
$-4\le4x\le-2$
Divide all three parts of the inequality by $4$:
$-\dfrac{4}{4}\le\dfrac{4x}{4}\le-\dfrac{2}{4}$
$-1\le x\le-\dfrac{1}{2}$
Expressing the solution in interval notation:
$\Big[-1,-\dfrac{1}{2}\Big]$