Solve an absolute value equation simultaneously My question is : 

Solve simultaneously
  $$\left\{\begin{align*}&|x-1|-|y-2|=1\\&y = 3-|x-1|\end{align*}\right.$$

What I did : 
$y=3 - |x-1|$ is given.
Thus $y = 3-(x-1)$ or $y = 3-\left(-(x-1)\right),$ and so $$y = 4-x\qquad\mbox{ or } \qquad y = 2+x.$$
If $y = 2+x$, then $$x - 1 = 4-y \quad(1)\qquad\mbox{ or } \qquad y - 2 = x \quad(2).$$
Substituting 1), we get $$|4-y| - |y-2| = 1  \qquad\mbox{ or } \qquad  |y-2-1|-|y-2| = 1.$$
I got here but I am not getting how to get the final solution. Any help would be greatly appreciated.

It’s entirely possible to solve this system by an exhaustive analysis of cases along the lines on which you’ve begun, but there’s an easier way. Notice that both equations have a term $|x-1|$. Let’s let $u=|x-1|$ and see whether the resulting system is a little easier to solve. After the substitution the original system 
$$\left\{\begin{align*}&|x-1|-|y-2|=1\\&y = 3-|x-1|\end{align*}\right.\tag{1}$$ becomes
$$\left\{\begin{align*}&u-|y-2|=1\\&y=3-u\end{align*}\right.\;.\tag{2}$$
The second equation of $(2)$ implies that $y-2=1-u$; substituting this into the first equation yields $$u-|1-u|=1\;.\tag{3}$$ If $u<1$, then $1-u>0$, and therefore $|1-u|=1-u$, and $(3)$ reduces to $u-(1-u)=1$, which is easily solved to get $u=1$ and hence $y=3-u=2$. Recall that $u=|x-1|$; thus, $|x-1|=1$, $x-1=\pm 1$, and $x=0$ or $x=2$. You can check that if $y=2$ and $x=0$ or $x=2$, then $(1)$ is satisfied.
If $u\ge1$, then $|1-u|=u-1$, and $(3)$ reduces to $u-(u-1)=1$, or $1=1$. This is always true, so every value of $u>1$ leads to a solution. Let’s check that. Let $u$ be any number greater than $1$, and let $y=3-u$. Then $y-2=1-u<0$, so $|y-2|=2-y=u-1$, and $$u-|y-2|=u-(2-y)=u-(u-1)=1\;,$$ so $(2)$ is indeed satisfied. All we need to do now is solve for $x$ in this case. We have $|x-1|=u$, so $x-1=\pm u$, and $x=1\pm u$. In other words, for each $u\ge 1$ there are two solutions,
$$\left\{\begin{align*}&x=1+u\\&y=3-u=3-(x-1)=4-x\end{align*}\right.\tag{4}$$ and
$$\left\{\begin{align*}&x=1-u\\&y=3-u=3-(1-x)=2+x\;.\end{align*}\right.\tag{5}$$
Finally, $(4)$ and $(5)$ can be simplified by getting rid of $u$. Since $u\ge 1$, $(4)$ gives us every value of $x\ge 2$, and $(5)$ gives us every value of $x\le 0$. Thus, $(4)$ and $(5)$ reduce to saying that the solutions to $(1)$ are 
$$\begin{align*}
&y=4-x\text{ for any }x\ge 2\\
&y=2+x\text{ for any }x\le 0\;,
\end{align*}\tag{6}$$
with no solutions having $0<x<2$. You can improve a little on $(6)$ by recasting it in terms of $y$: note that both branches of the solution give you all values of $y\le 2$, so you can rephrase $(6)$ by saying that the solutions are all pairs such that 
$$y\le 2,\text{ and }x=4-y\text{ or }x=y-2\;.$$
