3 Task 1. Solve the system of equations

$$
\left\{\begin{array}{l}
|y-x|=2(x+y) \\
|x|+|y|=4
\end{array}\right.
$$
Solution. The left side of equation (1) is non-negative. Therefore, $x+y \geqslant 0$, i.e., $y \geqslant -x$.

If $|y-x|=0$, i.e., $y=x$, then $x=0$.

If $y>x$, then equation (1) takes the form $y-$ $-x=2(x+y)$. From this, $y=-3x$.

If $y<x$, then from equation (1) we get $x-y=$ $=2(x+y)$ and $y=-x: 3$.

Thus, the graph of $f(x)$ of equation (1) is the union of rays $OA$ and $OB$ (Fig. 1).

Since

$$
\begin{aligned}
|x|+|y|= & |-x|+|y|=|-x|+|-y|= \\
& =|x|+|-y|
\end{aligned}
$$

the axes of symmetry of the graph of equation (2) are the x-axis and the y-axis. First, we construct that part of the graph $\varphi(x)$ of equation (2) which belongs to the first coordinate quadrant:

$$
x+y=4, x \geqslant 0, y \geqslant 0
$$

We obtain the segment $CD$. To construct the entire graph of $f(x)$, it is sufficient to reflect the segment $CD$ relative to the coordinate axes.

Now it is clear that the solutions to the given system of equations are the pairs of numbers $(-1 ; 3),(3,-1)$.