3 Task 2. Solve the system of equations

$$
\left\{\begin{array}{l}
|x|+x=|y|+y \\
|x-1|=x+|y|
\end{array}\right.
$$
Solution. First, let's investigate equation (1):

If $x \geqslant 0$ and $y \geqslant 0$, then $2 x=2 y$, i.e., in this case, the graph of equation (1) is the bisector OA of the first coordinate angle (Fig. 2).

If $x<0$ and $y \geqslant 0$, then $0=2 y$, i.e., in this case, the graph of equation (1) will be the ray $O C$.

Let $x \leqslant 0$ and $y \leqslant 0$. Then equation (1) takes the form $0=0$. This means that all points in the third coordinate angle are solutions to equation (1).

Fig. 1
![](https://cdn.mathpix.com/cropped/2024_05_21_8debcc8c8285c4d79a3dg-071.jpg?height=496&width=1030&top_left_y=254&top_left_x=525)

Fig. 2

Finally, let $x \geqslant 0, y \leqslant 0$. Then equation (1) transforms to $2 x=0$, i.e., $x=0$.

Thus, the solutions to equation (1) are all points in the third coordinate angle (including its sides) and the points on the bisector $O A$ of the first coordinate angle. Now, let's investigate equation (2):

If $x \geqslant 1$, then $x-1=x+|y|$, i.e., in this case, equation (2) has no solutions.

Let $x<1$. We get $1-x=x+|y|$ or $x=0.5(1-|y|)$.

Thus, the graph of equation (2) is the union of rays $D F$ and $D E$.

We find the coordinates of point $P$, where rays $O A$ and $D F$ intersect. Segment $O P$ is the bisector of angle $D O F$ of triangle $D O F$. Therefore, $D P: P F=O D: O F=1: 2$. From this, it follows that $P\left(\frac{1}{3} ; \frac{1}{3}\right)$.

Thus, the solution to the given system of equations (1) and (2) is each point on ray $E H$ and point $P\left(\frac{1}{3} ; \frac{1}{3}\right)$.