solving absolute value equation 2

My question is : Solve simultaneously-

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

I tried to solve this question by the method told by Marvis as I had understood that method (its here: Solve an absolute value equation simultaneously

But the solution set i got for the above question is not correct.
My solution was: $y \geq 2$, and x=3 or x=y-2.
I would like to know the final correct solution.

Answer

We shall proceed on similar lines as the answer here.

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 $y > 2$, then we get that $$y - y+ 2 =2,$$ which is true for all $y >2$.

If $y \leq 2$, then we get that $$y + y-2 =2 \implies y=2$$ Hence we get that $$y \geq 2$$ From the second equation, we get that $$\lvert x - 1 \rvert = 3 - y$$
Since $\lvert x - 1 \rvert \geq 0$, we need $$3-y \geq 0$$ This means that $y \leq 3$. Hence, we have that $2 \leq y \leq 3$.

If $x \geq 1$, then $x-1 = 3-y \implies x = 4-y$. Note that since $y \in [2,3]$, $x = 4-y \geq 1$.

If $x < 1$, then $x-1 = y-3 \implies x = y-2$. Note that since $y \in [2,3]$, $x = y-2 < 1$.

Hence, the solution set is given as follows. $$2 \leq y \leq 3 \\ \text{ and }\\ x = 4-y \text{ or } y-2$$