Answer
$\text{the interval } \left( -\infty,-\dfrac{25}{3} \right)\cup \left( \dfrac{35}{3},\infty \right)
$
Work Step by Step
Since $|x|\gt a$ implies $x\gt a \text{ OR } x\lt -a$, then the statement $
|\dfrac{3x-5}{6}|\gt5
$ evaluates to
\begin{array}{l}
\dfrac{3x-5}{6}\gt5\\\\
3x-5\gt30\\
3x\gt35\\\\
x\gt\dfrac{35}{3}
,\\\\\text{OR}\\\\
\dfrac{3x-5}{6}\lt-5\\\\
3x-5\lt-30\\
3x\lt-25\\\\
x\lt-\dfrac{25}{3}
.\end{array}
Hence, the solution set is $
\text{the interval } \left( -\infty,-\dfrac{25}{3} \right)\cup \left( \dfrac{35}{3},\infty \right)
$.
