Solve the following system of equations:

$$
x+\frac{x+2 y}{x^{2}+y^{2}}=2, \quad y+\frac{2 x-y}{x^{2}+y^{2}}=0
$$
Multiply the first equation by $y$, and the second by $x$, and add the resulting equations: then we surprisingly get the simple

$$
x y+1=y
$$

relation.

Now let's rearrange the second equation so that the substitution is as simple as possible:

$$
x(x y+2)+y^{3}-y=0
$$

On the left side, we apply the condition obtained from (1)

$$
x y=y-1
$$

twice, so we finally get

$$
y^{3}+x-1=0
$$

Using the relation also obtainable from (1) - by excluding the clearly impossible $y=0$ - 

$$
x=1-\frac{1}{y}
$$

equation, (2) takes the form

$$
y^{3}-\frac{1}{y}=0
$$

from which we immediately get that $y=1$ or $y=-1$. In the first case, according to (3), $x=0$, and in the second case $x=2$. (We are restricting ourselves to real numbers.)

Thus, the system of equations has two solutions:

$$
\begin{array}{ll}
x=0 ; & y=1 \\
x=2 ; & y=-1
\end{array}
$$