algebra precalculus - 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.

Answer

You have that $\lvert x - 1 \rvert - \lvert y - 2 \rvert = 1$. This gives us that

$$\lvert x - 1 \rvert = 1 + \lvert y-2 \rvert.$$ Plugging this into the second equation gives us $$y = 3 - \left( 1 + \lvert y-2 \rvert \right) = 2 - \lvert y - 2 \rvert$$
This gives us that $$y + \lvert y - 2 \rvert = 2.$$
If $y \geq 2$, then we get that $$y + y- 2 =2 \implies y = 2$$
If $y < 2$, then we get that $$y + 2 - y =2$$ which is true for all $y <2$. Hence we get that $$y \leq 2$$ From the second equation, we get that $$\lvert x - 1 \rvert = 3 - y$$
Note that since $y \leq 2$, $3-y > 0$ always.

If $x >1$, then

$$x - 1 = 3 - y \implies x = 4-y$$ Note that $4-y > 1$, since $y \leq 2$.

If $x \leq 1$, then

$$x - 1 = y-3 \implies x = y-2$$ Note that $y-2 \leq 1$, since $y \leq 2$.

Hence, the solution set is given as follows.

$$y \leq 2 \\ \text{ and }\\ x = 4-y \text{ or } y-2$$