<s>[INST] 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.
 [/INST] Whenever you see an absolute, you need to consider the possible cases:
$$|x-1|=\begin{cases} x-1 & x \geq 1 \\-(x-1) & x<1\\\end{cases}$$

As for the first equation, you might consider four cases:


*

*$y \geq 2$ and $x \geq 1$:
$$(x-1)-(y-2)=1 \implies y=x$$
$$y=3-(x-1) \implies y=4-x$$
$$\Downarrow$$
$$\left\{\begin{align*}&y=x\\&y=4-x\end{align*}\right.$$

*$y \geq 2$ and $x <1$:
$$-(x-1)-(y-2)=1$$
$$y=3+(x-1)$$
$$\Downarrow$$
$$\left\{\begin{align*}&y=2-x\\&y=x+2\end{align*}\right.$$


*

*$y<2$ and $x \geq 1$:
$$(x-1)+(y-2)=1$$
$$y=3-(x-1) $$

*$y<2$ and $x <1$:


$$-(x-1)+(y-2)=1$$
$$y=3+x-1$$
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