For absolute value equalities, say $|x-a|=b$, my approach is :
There are two cases, based on which the sign of the absolute value is taken.
Case 1: If $x-a\ge b$, then $x-a\ge 0$; so for given interval $x\ge a+b$, $x = a+b$.
Case 2: If $x-a\lt b$ then $x-a\lt 0$; so for given interval $x\lt a+b$, $x = a-b$.
But, have two posts on mse that contradict.
One is an answer to my post, and other is here.
For the first example, have reverse inclusion, i.e. left end is excluded; while the right end is included. This has yielded an extra solution point. However, substitution of the extra solution $x=2$ yields a correct value on substitution back, as follows:
$4=4$, which is correct.
It is logical too, as in the next interval all of the points on real axis for which $x\ge 2$ are valid.
But, still it is a logical error as per me.
The other post, has error (as per me) in the second case where both left and right end are included.
Thanks to @SiongThyeGoh for pointing out errors in my solution construction.
Have not specified (i) $b\ge 0$,
(ii) Should specify for the two cases instead as :
Case 1: If $x-a\ge 0$, then $x-a=b\implies x = a+b$,
Case 2: If $x-a\lt 0$, then $x-a=-b\implies x = a-b$,