6.185. $\left\{\begin{array}{l}x^{2}+2 y^{2}=17 \\ x^{2}-2 x y=-3\end{array}\right.$

The system of equations is:
\[
\left\{\begin{array}{l}
x^{2}+2 y^{2}=17 \\
x^{2}-2 x y=-3
\end{array}\right.
\]
## Solution.

Since the left sides of the system are homogeneous, we make the substitution $y = t x$. Then the system will take the form $\left\{\begin{array}{l}x^{2}\left(1+2 t^{2}\right)=17, \\ x^{2}(1-2 t)=-3 .\end{array}\right.$

By dividing the left and right parts of the equations $(x \neq 0)$, we get

$$
\frac{1+2 t^{2}}{1-2 t}=-\frac{17}{3}\left(t \neq \frac{1}{2}\right) \Leftrightarrow 3 t^{2}-17 t+10=0, t_{1}=\frac{2}{3}, t_{2}=5
$$

1) $t_{1}=\frac{2}{3}$. From the equation $x^{2}(1-2 t)=-3$ we have $x^{2}\left(1-\frac{4}{3}\right)=-3, x^{2}=9$, $x_{1}=3, x_{2}=-3 ; y_{1}=\frac{2}{3} \cdot 3=2, y_{2}=\frac{2}{3}(-3)=-2$;
2) $t_{2}=5$. Then $x^{2}(1-10)=-3, x^{2}=\frac{1}{3}, x_{1}=-\frac{1}{\sqrt{3}}, x_{2}=\frac{1}{\sqrt{3}} ; y_{1}=5\left(-\frac{1}{\sqrt{3}}\right)=$ $=-\frac{5}{\sqrt{3}}, y_{2}=5 \cdot \frac{1}{\sqrt{3}}=\frac{5}{\sqrt{3}}$.

Answer: $(3 ; 2),(-3 ;-2),\left(\frac{\sqrt{3}}{3} ; \frac{5 \sqrt{3}}{3}\right),\left(-\frac{\sqrt{3}}{3} ; \frac{-5 \sqrt{3}}{3}\right)$.