Linear equation with absolute value. [closed]

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

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

closed as off-topic by user21820, Lee David Chung Lin, Ernie060, Arnaud D., José Carlos SantosApr 29 at 13:34

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• What does obsolete mean for a value ? – Rebellos Apr 15 at 19:02
• They do not satisfy the equation – user664431 Apr 15 at 19:03
• I dunno what they don't – user664431 Apr 15 at 19:03

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?