Solve the following system of equations:

$$
\begin{gathered}
3(x+y)+2\left(x^{2}+y^{2}\right)=41 \\
2 x y+3\left(x^{2}+y^{2}\right)=51
\end{gathered}
$$
Subtracting the first equation from the second

$$
x^{2}+2 x y+y^{2}-3(x+y)=10
$$

Rearranging and substituting $x+y=u$

$$
u^{2}-3 u-10=0
$$

from which

(1)

$$
u_{1}=5, u_{2}=-2
$$

Subtracting the second equation from twice the first equation

$$
x^{2}-2 x y+y^{2}+6 u=31
$$

Let $x-y=v$, then

$$
v^{2}=31-6 u
$$

Substituting the values of $u$ into this equation

$$
v_{1}^{2}=1, v_{2}^{2}=43
$$

and thus

$$
v_{1}= \pm 1, v_{2}= \pm \sqrt{43}
$$

From (1) and (2)

$$
\begin{aligned}
& x_{1}=3, x_{2}=2, x_{3}=\frac{-2+\sqrt{43}}{2}, x_{4}=\frac{-2-\sqrt{43}}{2} \\
& y_{1}=2, y_{2}=3, y_{3}=\frac{-2-\sqrt{43}}{2}, y_{4}=\frac{-2+\sqrt{43}}{2}
\end{aligned}
$$

(Marczell Riesz, Györ.)

Number of solutions: 51.