How to solve equations involving modulus function of the type $|x+1| - |1-x|=2 $ and $ |x-1|=|x|+a$? I am able to solve equation of the type $ |5x+1|=|11-2x|$. I square both the side and my equation becomes $ (5x+1)^2=(11-2x)^2 $ further simplification gives me $ (5x+1)=\pm (11-2x)$.
I get  have difficulties in solving equation of the type 
$|x+1| - |1-x|=2~~$   and
$~~|x-1|=|x|+a$
Any steps and or links that could be of help will be greatly appreciated.

In this case, $|x + 1| - |1 - x| = 2$, we have,
$$
|x + 1| = \begin{cases} x+1 \ \text{if} \ x \geq -1\\ -x-1 \ \text{if} \ x < -1 \end{cases}
\quad \text{and} \quad
|1 - x| = \begin{cases} 1 - x \ \text{if} \ x \leq 1\\ x-1 \ \text{if} \ x \geq 1 \end{cases}
$$
If $x < -1, \ |x + 1| =-x-1$ and $|1 - x| = 1 - x$, then
$$
|x + 1| - |1 - x| = 2 \quad \Rightarrow \quad -x-1-(1 - x) = 2 \ \Rightarrow \ -2 = 2 \quad \text{Absurd!}
$$ 
If $-1 \leq x < 1, \ |x + 1| =x + 1$ and $|1 - x| = 1 - x$, then
$$
|x + 1| - |1 - x| = 2 \quad \Rightarrow \quad x + 1 - (1 - x) = 2 \ \Rightarrow \ 2x = 2 \quad \Rightarrow \quad x = 1 
$$
If $x > 1, \ |x + 1| = x + 1$ and $|1 - x| = x - 1$, then
$$
|x + 1| - |1 - x| = 2 \quad \Rightarrow \quad x + 1 - (x - 1) = 2 \ \Rightarrow \ 2 = 2 \quad \text{valid for all} \quad x > 1 
$$
Thus, $S = \{x \in \mathbb{R} : x \geq 1\}$. The other issue resolves also analyzing cases.
