Answer
$\left[ -\dfrac{1}{2},\dfrac{1}{2} \right]$
Work Step by Step
For any $a\gt0$, $|x|\lt a$ implies $-a\lt x\lt a$. (The symbol $\lt$ may be replaced with $\le$.)
Using the concept above, the solutions to the given inequality, $
|16x|\le8
,$ is
\begin{array}{l}\require{cancel}
-8\le 16x\le8
\\\\
-\dfrac{8}{16}\le x\le\dfrac{8}{16}
\\\\
-\dfrac{1}{2}\le x\le\dfrac{1}{2}
.\end{array}
In interval notation, the solution set is $
\left[ -\dfrac{1}{2},\dfrac{1}{2} \right]
.$
