The equation implies $x - |3x+1| = \pm 4$. Two cases.
If $3x+1 > 0$ then you have
$$
\begin{split}
x - (3x+1) &\in \{\pm 4\}\\
-2x &\in \{-3,5\} \\
x &\in \{3/2, -5/2\}
\end{split}
$$
but only one of those satisfies the original assumption $3x+1>0 \iff x > -1/3$, so we end up with a solution $x = 3/2$.
Alternatively, $3x+1 \le 0 \iff x \le -1/3$, which implies
$$
\begin{split}
x + (3x+1) &\in \{\pm 4\}\\
4x &\in \{3,-5\}\\
x &\in \{3/4, -5/4\}
\end{split}
$$
Can you finish the problem?
The equation implies $x - |3x+1| = \pm 4$. Two cases.
If $3x+1 > 0$ then you have $$ \begin{split} x - (3x+1) &\in \{\pm 4\}\\ -2x &\in \{-3,5\} \\ x &\in \{3/2, -5/2\} \end{split} $$ but only one of those satisfies the original assumption $3x+1>0 \iff x > -1/3$, so we end up with a solution $x = 3/2$.
Alternatively, $3x+1 \le 0 \iff x \le -1/3$, which implies $$ \begin{split} x + (3x+1) &\in \{\pm 4\}\\ 4x &\in \{3,-5\}\\ x &\in \{3/4, -5/4\} \end{split} $$ Can you finish the problem?