Answer
$x=\left\{ \dfrac{2}{3}, 1\right\}$
Work Step by Step
Isolating the absolute value expression, the given expression, $
|6x-5|-3=-2
,$ is equivalent to
\begin{array}{l}\require{cancel}
|6x-5|=-2+3
\\\\
|6x-5|=1
.\end{array}
Since for any $a$ greater than zero, $|x|=a$ implies $x=a$ OR $x=-a$, then
\begin{array}{l}\require{cancel}
6x-5=1
\\\\
6x=1+5
\\\\
6x=6
\\\\
x=\dfrac{6}{6}
\\\\
x=1
,\\\\\text{OR}\\\\
6x-5=-1
\\\\
6x=-1+5
\\\\
6x=4
\\\\
x=\dfrac{4}{6}
\\\\
x=\dfrac{\cancel{2}\cdot2}{\cancel{2}\cdot3}
\\\\
x=\dfrac{2}{3}
.\end{array}
Hence, $
x=\left\{ \dfrac{2}{3}, 1\right\}
.$
