<s>[INST] How could we solve $x$, in $|x+1|-|1-x|=2$? How could we solve $x$, in $|x+1|-|1-x|=2$? 
Please suggest a analytical way that I could use in other problems too like this $ |x+1|+|1-x|=2$ and of this genre.
Thank you,
 [/INST] (This isn’t really an answer, but it’s too big for a comment. It’s just continuing from where Nana stopped in their answer, revealing that you can’t find the solution using that approach.)
We have just squared both sides to try to get rid of the square roots.
$$ \left( \sqrt{(x+1)^2} \right)^2 = \left( 2 + \sqrt{(1-x)^2} \right)^2 $$
Continuing from here:
Cancel on the left, expand the square on the right.
$$ (x+1)^2 = 4 + 4 \sqrt{(1-x)^2} + (1-x)^2 $$
Expand squares.
$$ x^2 + 2x + 1 = 4 + 4 \sqrt{(1-x)^2} + 1 - 2x + x^2 $$
Combine terms.
$$ 4x - 4 = 4 \sqrt{(1-x)^2} $$
Divide by coefficient.
$$ x - 1 = \sqrt{(1-x)^2} $$
Square both sides again to get rid of that last square root.
$$ (x - 1)^2 = \left(\sqrt{(1-x)^2}\right)^2 $$
Cancel on the right.
$$ (x - 1)^2 = (1-x)^2 $$
Expand squares.
$$ x^2 - 2x + 1 = 1 - 2x + x^2 $$
Rearrange terms.
$$ x^2 - 2x + 1 = x^2 - 2x + 1 $$
Combine terms.
$$ 0 = 0 $$
The equation $0=0$ doesn’t help us find the correct values of $x$. To solve the problem, we must use another approach.
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