## Task 2 - 241222

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

$$
\begin{aligned}
x^{2} y^{2}+x^{2}+y^{2}-6 & =9 x y \\
(x+y)^{2} & =36
\end{aligned}
$$
We substitute $x+y=a$ and $x y=b$:

$$
b^{2}+a^{2}-2 b-6=9 b
$$

and

$$
a^{2}=36
$$

Substituting the second into the first equation:

$$
b^{2}-11 b+30=0
$$

Using Vieta's formulas, we guess $b_{1}=5$ and $b_{2}=6$, and from above $a_{1}=6$ and $a_{2}=-6$. Generally, for $x$ and $y$:

$$
y=\frac{b}{x}
$$

$$
\begin{gathered}
x+\frac{b}{x}=a \\
x^{2}-a x+b=0 \\
x=\frac{a}{2} \pm \sqrt{\frac{a^{2}}{4}-b}
\end{gathered}
$$

We now need to combine each $a_{i}$ with each $b_{i}$ to find all solutions. $a_{1}$ and $b_{1}$:

$$
x_{1,2}=3 \pm \sqrt{9-5}=3 \pm 2
$$

Solution pairs: $(5,1)$ and $(1,5)$. $a_{1}$ and $b_{2}$:

$$
x_{3,4}=3 \pm \sqrt{9-6}=3 \pm \sqrt{3}
$$

Solution pairs: $(3+\sqrt{3}, 3-\sqrt{3})$ and $(3-\sqrt{3}, 3+\sqrt{3})$. $a_{2}$ and $b_{1}$:

$$
x_{5,6}=-3 \pm \sqrt{9-5}=-3 \pm 2
$$

Solution pairs: $(-5,-1)$ and $(-1,-5)$. $a_{2}$ and $b_{2}$:

$$
x_{7,8}=-3 \pm \sqrt{9-6}=-3 \pm \sqrt{3}
$$

Solution pairs: $(-3+\sqrt{3},-3-\sqrt{3})$ and $(-3-\sqrt{3},-3+\sqrt{3})$. Thus, there are 8 solution pairs.