324. Solve the system of equations:

$$
\left\{\begin{array}{l}
x^{2}+x y+y=1 \\
y^{2}+x y+x=1
\end{array}\right.
$$
$\triangle$ Subtract the equations of the system:

$$
x^{2}+y-y^{2}-x=0, \quad(x-y)(x+y-1)=0
$$

Now we need to consider two cases: $x-y=0$ and $x+y-1=0$. Work through these cases on your own.

Answer: $(-1 ;-1),(x ; 1-x)$, where $x-$ is any number.