Tricky double absolute value equation For the equation
$$|2x − 1| − |x + 2| = 5$$
we want to find the solution set $S$.


*

*My own way: absolute expression #$1$ has two cases, and absolute expression #$2$ has two cases, so $2{\,\cdot\,} 2=4$ cases, which gives $S=\{-2,4/3,8\}$.

*But the answer to problem $41$ in this solution booklet
has $S=\{-2,8\}$.
So which one is right?
Thanks a lot!
 A: The graphical approach in the document you referenced works well.

The same argument, in algebraic form, goes like this . . .

Let $f(x)= |2x - 1| - |x + 2|$.



*

*On the interval $(-\infty,-2)$, we get 
$$f(x)=(1-2x)-(-x-2)=-x+3$$
but the equation $-x+3=5$ yields $x=-2$, which is not in the interval $(-\infty,-2)$.
$\\[8pt]$

*On the interval $[-2,\frac{1}{2})$, we get 
$$f(x)=(1-2x)-(x+2)=-3x-1$$
and the equation $-3x-1=5$ yields $x=-2$, which qualifies as a solution, since $-2$ is in the interval $[-2,\frac{1}{2})$.
$\\[8pt]$

*On the interval $[\frac{1}{2},\infty)$, we get 
$$f(x)=(2x-1)-(x+2)=x-3$$
and the equation $x-3=5$ yields $x=8$, which qualifies as a solution, since $8$ is in the interval $[\frac{1}{2},\infty)$.


Hence there are two solutions: $x=-2,\;x=8$.

As regards the error in your reasoning . . .

Your approach was to interpret the solutions of the equation
$$|2x - 1| - |x + 2|=5\tag{*}$$
as the solutions of any of the $4$ equations
$$\pm(2x-1)-\left(\pm(x+2)\right)=5\tag{**}$$
Of course, any solution to $(*)$ must be a solution to one of the equations from $(**)$.

However, the converse need not be true.

For example, in this case, the equation
$$+(2x-1)-\left(-(x+2)\right)=5$$
is an instance of $(**)$, but does not qualify as a case for $(*)$, since 


*

*The reduction $|2x-1|=+(2x-1)$ holds for $x\ge {\large{\frac{1}{2}}}$.$\\[4pt]$

*The reduction $|x+2|=-(x+2)$ holds for $x\le -2$.


but you can't have both $x\ge {\large{\frac{1}{2}}}$ and $x\le -2$.

Hence, if you choose to use your approach without worrying about which cases might not be possible, you need to test your solutions, and reject those which don't satisfy the original equation. For the case at hand, you found $3$ solutions, and of those, only $x=-2$ and $x=8$ satisfy the original equation.
