## Task 1 - 251221

Determine all pairs $(x ; y)$ of real numbers $x, y$ that satisfy the following system of equations (1), (2).

$$
\begin{aligned}
& x^{2}+y^{2}=5 \\
& x^{2}+x y=2
\end{aligned}
$$
Adding twice equation (2) to equation (1) yields

$$
9=3 x^{2}+2 x y+y^{2}=2 x^{2}+(x+y)^{2}
$$

From the second equation, it follows on the one hand that $x \neq 0$, since otherwise a contradiction $0=2$ would arise, and on the other hand, this implies $x+y=\frac{2}{x}$. Substituting this into the equation just obtained and multiplying by $x^{2}$, we get

$$
9 x^{2}=2 x^{4}+4 \quad \text { or } \quad 2 z^{2}-9 z+4=0
$$

with $z:=x^{2}$. The solutions to this quadratic equation in $z$ are $\frac{9}{4} \pm \sqrt{\frac{81}{16}-2}=\frac{9 \pm \sqrt{81-32}}{4}=\frac{9 \pm 7}{4}$, i.e., $z_{1}=\frac{1}{2}$ and $z_{2}=2$.

For $z_{1}=\frac{1}{2}$, the values are $x_{1 / 2}= \pm \frac{1}{2} \cdot \sqrt{2}$, and thus from equation (2), $y_{1 / 2}= \pm \frac{3}{2} \cdot \sqrt{2}$, where the same sign must be chosen in each case. That these solutions also satisfy the first equation can be easily verified.

On the other hand, from $x^{2}=z_{2}=2$, the second equation would yield $y=0$, which, however, contradicts the first equation.

Thus, there are exactly two solutions, namely $(x, y) \in\left\{\left(\frac{1}{2} \cdot \sqrt{2}, \frac{3}{2} \cdot \sqrt{2}\right) ;\left(-\frac{1}{2} \cdot \sqrt{2},-\frac{3}{2} \cdot \sqrt{2}\right)\right\}$.