9.2. Solve the system of equations

$$
\left\{\begin{aligned}
10 x^{2}+5 y^{2}-2 x y-38 x-6 y+41 & =0 \\
3 x^{2}-2 y^{2}+5 x y-17 x-6 y+20 & =0
\end{aligned}\right.
$$
Solution: We will eliminate the product $xy$. For example, multiply the first equation by 5, the second by 2, and add the left and right parts of the obtained equations. We get $56x^2 + 21y^2 - 224x - 42y + 245 = 0$. Divide the equation by 7, and then complete the squares for $x$ and $y$. We have $8(x-2)^2 + 3(y-1)^2 = 0$, from which $x=2, y=1$. By verification, we can ensure that the found pair of numbers is indeed a solution.

Answer: $x=2, y=1$.

Recommendations for checking:

| is in the work | points |
| :--- | :--- |
| Correct and fully justified answer | 7 points |
| The pair $x=2$ is justified (other pairs $(x, y)$ are not solutions), but there is no check that it solves the system | 4 points |
| The equation not containing the product $xy$ is correctly obtained by valid transformations; no further progress | 2 points |
| Answer without justification or with a check that the pair $x=2$, $y=1$ is a solution | 1 point |
| Transformations not leading to a solution | 0 points |