5. Find all solutions of the system of equations

$$
\begin{aligned}
x^{2}+y^{2} & =25 \text { and } \\
x+y+x y & =19
\end{aligned}
$$

Solve the problems independently. You have 150 minutes for solving.

The use of notes, literature, or a pocket calculator is not allowed.

## Solutions to the selection competition problems

```
Each problem is worth 7 points. When evaluating, reasonably consider the provided scoring guide.
    A competitor should not receive more than 3 points for a particular problem if the partial solution does not clearly lead to the final solution of the problem.
```
IV/5. Multiply the second equation by 2 and add the first

$$
x^{2}+y^{2}+2 x+2 y+2 x y=25+2 \cdot 19
$$

we transform it into $(x+y)^{2}+2(x+y)=63$ and factorize $(x+y-7)(x+y+9)=0$. If $x+y=7$, from the second equation it follows that $x y=12$. From these equations, we get a quadratic equation for $x$, which factors nicely into $(x-3)(x-4)=0$. We obtain the solutions $x=3, y=4$ and $x=4, y=3$. For $x+y=-9$, it follows that $x y=28$, and the quadratic equation for $x$ that we derive, i.e., $x^{2}+9 x+28$, has a negative discriminant, so there are no other solutions.

Transforming into $(x+y-7)(x+y+9)=0$ (or equivalently, which gives linear relations between $x$ and $y$): $\mathbf{3}$ points. The first quadratic equation $(x-3)(x-4)=0$ and both solutions: $\mathbf{1 + 1}$ point. The second quadratic equation $x^{2}+9 x+28$ and proof that there are no real solutions: $\mathbf{1 + 1}$ point.

If the contestant does not solve the problem but expresses $y$ (or $x$) from the second equation and writes the correct polynomial of degree 4 in $x$ (or $y$): 3 points.