2. For which values of the parameter $a$ does the system of equations $|x-1|+|y-a|=1$ and $y=-2|x-1|-1$ have exactly 3 solutions?
I/2. We draw curves defined by equations. The first equation defines the perimeter of a square centered at the point $(1, a)$, the second defines 2 half-rays with a common origin at the point $(1,-1)$. The system has exactly 3 solutions when the common origin of the half-rays coincides with the upper vertex of the square. This is the case when $a=-2$.

![](https://cdn.mathpix.com/cropped/2024_06_07_5a4348e8f3c4edef190ag-06.jpg?height=314&width=312&top_left_y=1182&top_left_x=1563)

Determination that the first equation defines the perimeter of a square 2 points

Determination of the meaning of the parameter $a$ 1 point

Determination that the second equation defines half-rays with a common origin at $(1,-1)$ 2 points

Determination of the condition for exactly 3 solutions 1 point

Solution $a=-2$

1 point