6.184. $\left\{\begin{array}{l}2 x^{2}-3 x y+y^{2}=3 \\ x^{2}+2 x y-2 y^{2}=6\end{array}\right.$
## Solution.

Since the left-hand sides of the equations are homogeneous with respect to $x$ and $y$, let's introduce the substitution $y = t x$; the system of equations will take the form

$$
\left\{\begin{array}{l}
x^{2}\left(2-3 t+t^{2}\right)=3 \\
x^{2}\left(1+2 t-2 t^{2}\right)=6
\end{array}\right.
$$

Dividing the left and right parts of the equations $(x \neq 0)$, we get $\frac{2-3 t+t^{2}}{1+2 t-2 t^{2}}=\frac{1}{2}\left(t \neq \frac{1 \pm \sqrt{3}}{2}\right), 4 t^{2}-8 t+3=0$, from which $t_{1}=\frac{1}{2}, t_{2}=\frac{3}{2} ;$

1) $t_{1}=\frac{1}{2}$. From the equation $x^{2}\left(1+2 t-2 t^{2}\right)=6$ we find $x^{2}\left(1+1-\frac{1}{2}\right)=6$, $x^{2}=4$, from which $x_{1}=2, x_{2}=-2 ;$ then $y_{1}=1, y_{2}=-1$;
2) $t_{2}=\frac{3}{2}$. Then $x^{2}\left(1+3-\frac{9}{2}\right)=6, x^{2}=-12$, no solutions exist.

Answer: $(2 ; 1),(-2 ;-1)$.