Linear equation with absolute value.

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Find $x$ if $|x-|3x+1||=4$

I got 4 values of $x$ out of which 2 are obsolete... Why so??

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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?