Determine solutions for an absolute value equation

Question

I have 2 euqations that looks very similar, and here are my stpes：

First one: $$ \begin{eqnarray} 3|x-1|-2 & = & 10 \\ 3|x-1| & = & 12 \\ |x-1| & = & 4 \\ x-1 & = & \pm4 \end{eqnarray} $$
Second one: $$ \begin{eqnarray} |x+1|-5 & = & 3x \\ |x+1| & = & 3x+5 \\ x+1 & = & \pm(3x+5) \end{eqnarray} $$
I was told that my second solution was wrong becasue there was supposed to be only one solution for the second equation. What am I getting wrong?

Answer 1

$|x|=y$ if and only if $x= y $ or $x=-y$. We have to check which one (or both)holds.

In this case, from $|x+1|=3x+5$, we require that $3x+5\ge 0$.

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$x+1=3x+5$ or $x+1=-3x-5$.

$2x=-4$ or $4x=-6$

$x=-2$ or $x=-\frac32$.

We can reject the first solution by checking $3x+5\ge 0$ is violated.