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.
 A: For the first one,
$$|x+1| - |1-x|=2$$
Or,
$$|x+1| - |x-1|=2$$
as $|x|=|-x|$, I just rearranged just because I prefer x first, not necessary;
Now break a number line into three parts:
Use:
$|x|=\begin{cases}+x,\;x>0\\-x,\;x<0\end{cases}$ 
Part I:$\;x\in(-\infty,-1]$
Now, $$x+1\le0\wedge x-1<0$$
So, $$-(x+1)+(x-1)=-2$$ no solution here;
Part II:$\;x\in(-1,1]$
Now, $$x+1>0\wedge x-1\le0$$
So, $$(x+1)+(x-1)=2\implies x=1$$which is in this range;
Part III:$\;x\in(1,\infty)$
Now, $$x+1>0\wedge x-1>0$$
So, $$+(x+1)-(x-1)=2$$which is true for all x in the required range;

So, $$\large\boxed{x\in[1,\infty)}$$

Solve second one similiarly by breaking into $(\infty,0],(0,1],(1,\infty)$

Note: There is a simpler way way for first, which is a special case:
Consider it as:
$$|x-(-1)|-|x-(1)|=(1)-(-1)$$
Let P(x),A(-1) and B(1) be points on number line, then:
$$PA-PB=AB$$ And it is clear from the following figure that $x\in[1,\infty)$

A: You could use two steps of the same kind you showed before:
$|x+1|=2+|1-x|$
$x+1=2+|1-x|$ or $x+1=-2-|1-x|$
$x-1=|1-x|$ or $-x-3=|1-x|$
$x-1=1-x$ or $x-1=-1+x$ or $-x-3=1-x$ or $-x-3=-1+x$
and then solve each of these four equations seperately.
A: As explained in some answers to this question and also to How could we solve $x$, in $|x+1|-|1-x|=2$?, you can approach this geometrically.
So for the second equation $$|x-1|-|x|=a$$ you are looking for the points such that the difference between the distance from $1$ and distance from $0$ is equal to $a$.
You can also draw the graphs of the function $|x-1|$, $|x|$ and of their difference; this might help your insight into the problem.
Here are the graphs drawn by WolframAlpha.
We can see (either from graph or from geometric interpretation) that:


*

*If $|a|>1$, then there are no solutions. (See also reverse triangle inequality, which implies that $|a|=||x-1|-|x||\le 1$.) 

*If $a=1$, then solutions are all numbers from the interval $(-\infty,0]$.

*If $a=-1$, then solutions are all numbers from the interval $[1,\infty)$.

*If $a\in(-1,1)$, then there will be exactly one solution somewhere between $0$ and $1$. I leave the details to you, but you should be able to show that $x=\frac{1-a}2$.

A: 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.
