Answer
$\text{the interval } \left( -\infty,0 \right]\cup \left[ \dfrac{6}{5},\infty \right)
$
Work Step by Step
Since $|x|\gt a$ implies $x\gt a \text{ OR } x\lt -a$, then the statement $
8+|5x-3|\ge11
$ evaluates to
\begin{array}{l}
|5x-3|\ge3\\
5x-3\ge3\\
5x\ge6\\\\
x\ge\dfrac{6}{5}
,\\\\\text{OR}\\\\
5x-3\le-3\\
5x\le0\\
x\le0
.\end{array}
Hence, the solution set is $
\text{the interval } \left( -\infty,0 \right]\cup \left[ \dfrac{6}{5},\infty \right)
$.
