Answer
the interval $\left(-\infty, -2 \right)\cup\left(\dfrac{4}{3},\infty\right)$
Work Step by Step
Since for any $a\gt0$, $|x|\gt a$ implies $x\gt a$ or $x\lt-a$ the solution to the given inequality, $
|3x+1|\gt5
,$ is
\begin{array}{l}\require{cancel}
3x+1\gt5
\\\\
3x\gt5-1
\\\\
3x\gt4
\\\\
x\gt\dfrac{4}{3}
,\\\\\text{OR}\\\\
3x+1\lt-5
\\\\
3x\lt-5-1
\\\\
3x\lt-6
\\\\
x\lt-\dfrac{6}{3}
\\\\
x\lt-2
.\end{array}
Hence, the solution set is $\text{
the interval $\left(-\infty, -2 \right)\cup\left(\dfrac{4}{3},\infty\right)$
.}$
