Answer
$\left( -\infty,\dfrac{6}{5}
\right]\cup\left[ 2,\infty \right)
.$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To solve the given equation, $
|8-5x|\ge2
,$ use the definition of absolute value inequalities.
$\bf{\text{Solution Details:}}$
Since for any $c\gt0$, $|x|\gt c$ implies $x\gt c \text{ or } x\lt-c$ (which is equivalent to $|x|\ge c$ implies $x\ge c \text{ or } x\le-c$), the inequality above is equivalent to
\begin{array}{l}\require{cancel}
8-5x\ge2
\\\\\text{OR}\\\\
8-5x\le-2
.\end{array}
Solving each inequality results to
\begin{array}{l}\require{cancel}
8-5x\ge2
\\\\
-5x\ge2-8
\\\\
-5x\ge-6
\\\\\text{OR}\\\\
8-5x\le-2
\\\\
-5x\le-2-8
\\\\
-5x\le-10
.\end{array}
Dividing both sides by a negative number and consequently reversing the inequality symbol results to
\begin{array}{l}\require{cancel}
x\le\dfrac{-6}{-5}
\\\\
x\le\dfrac{6}{5}
\\\\\text{OR}\\\\
x\ge\dfrac{-10}{-5}
\\\\
x\ge2
.\end{array}
Hence, the solution set is the interval $
\left( -\infty,\dfrac{6}{5}
\right]\cup\left[ 2,\infty \right)
.$