5. a) Solve the system of equations

$$
|x|+|y-1|=1, x+2 y=3
$$

b) Graphically represent the solution in the plane of a rectangular coordinate system.
Solution. Let's consider the following four systems:

1) $x \geq 0, y \geq 1, x+y=2, x+2 y=3$,
2) $x \geq 0, y<1, x-y=0, x+2 y=3$,
3) $x<0, y \geq 1, y-x=2, x+2 y=3$,
4) $x<0, y<1, x+y=0, x+2 y=3$.

System 1) has a solution (1,1), systems 2) and

![](https://cdn.mathpix.com/cropped/2024_06_05_8b1793bf5544e167118bg-02.jpg?height=456&width=475&top_left_y=1750&top_left_x=993)

3) have no solution, while system 4) has a solution ( $-\frac{1}{3}, \frac{5}{3}$ ). The solutions to the system are shown in the diagram above. Specifically, the set of points ( $x, y$ ) for which $|x|+|y-1|=1$ is the boundary of the square with vertices $(0,0),(1,1),(0,2),(-1,1)$. The line $x+2 y=3$ intersects this boundary at the points (1,1) and ( $-\frac{1}{3}, \frac{5}{3}$ ).

## Second Year