2. (15 points) Solve the system of equations: $\left\{\begin{array}{l}y^{2}+x y=15 \\ x^{2}+x y=10\end{array}\right.$
Solution: By adding the equations, we get: $(x+y)^{2}=25$ or $x+y= \pm 5$.

Therefore, the system of equations splits into two:

$\left[\begin{array}{l}\left\{\begin{array}{l}x+y=5 \\ x^{2}+x y=10\end{array}\right. \\ \left\{\begin{array}{l}x+y=-5 \\ x^{2}+x y=10\end{array}\right.\end{array} \Leftrightarrow\left[\begin{array}{l}\left\{\begin{array}{l}y=5-x \\ x^{2}+x(5-x)=10\end{array}\right. \\ \left\{\begin{array}{l}y=-5-x \\ x^{2}+x(-5-x)=10\end{array}\right.\end{array} \Leftrightarrow\left[\begin{array}{l}\left\{\begin{array}{l}y=5-x \\ 5 x=10\end{array}\right. \\ \left\{\begin{array}{l}y=-5-x \\ -5 x=10\end{array}\right.\end{array} \Leftrightarrow\left[\begin{array}{l}\left\{\begin{array}{l}y=3 \\ x=2\end{array}\right. \\ \left\{\begin{array}{l}y=-3 \\ x=-2\end{array}\right.\end{array}\right.\right.\right.\right.$

Answer: $(2 ; 3),(-2 ;-3)$.

## Criteria

| Points | Conditions for awarding |
| :---: | :--- |
| 15 | Correct answer obtained with justification |
| 10 | Incorrect answer due to an arithmetic error |
| 0 | The solution does not meet any of the above conditions |