Answer
$(-\infty,-3] \cup [9,\infty)$
Work Step by Step
Using the properties of inequality, the given equation, $
\left| \dfrac{x}{3}-1 \right|-7\ge-5
,$ is equivalent to
\begin{array}{l}\require{cancel}
\left| \dfrac{x}{3}-1 \right|\ge-5+7
\\\\
\left| \dfrac{x}{3}-1 \right|\ge2
.\end{array}
Since for any $a>0$, $|x|\ge a$ implies $x\le-a$ or $x\ge a$, then the inequality above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{x}{3}-1\ge2
\\\\
x-3\ge6
\\\\
x\ge6+3
\\\\
x\ge9
,\\\\\text{OR}\\\\
\dfrac{x}{3}-1\le-2
\\\\
x-3\le-6
\\\\
x\le-6+3
\\\\
x\le-3
.\end{array}
Hence, the solution set is $
(-\infty,-3] \cup [9,\infty)
.$
