To solve the following system of equations:

$$
\begin{aligned}
& x + xy + y + 5 = 0 \\
& x^2 y + x y^2 + 6 = 0
\end{aligned}
$$
I. Solution: From the first equation

$$
y(x+1)+x+5=0
$$

Here, $x$ cannot be -1, so we can express $y$ as:

$$
y=-\frac{x+5}{x+1}
$$

Substituting this into the second equation and rearranging, we get the equation

$$
x^{4}+5 x^{3}-11 x^{2}-37 x-6=0
$$

Let's examine whether this equation has any rational roots. We know that if there is such a solution, its numerator must be a divisor of the constant term, -6, and its denominator must be a divisor of the coefficient of the highest degree term, which in this case is 1. Therefore, only the values $x= \pm 1, \pm 2, \pm 3$ and $\pm 6$ can be considered. Testing these, we find that -2 and 3 are indeed roots. Thus, the left side of the equation can be factored by the product of the corresponding root factors,

$$
(x+2)(x-3)=x^{2}-x-6
$$

and indeed, the left side of our fourth-degree equation can be written as:

$$
\left(x^{2}-x-6\right)\left(x^{2}+6 x+1\right)
$$

Thus, in addition to the roots we have already found, the roots of the equation are also the roots of the equation

$$
x^{2}+6 x+1=0
$$

The complete set of roots is therefore

$$
x_{1}=-2, \quad x_{2}=3, \quad x_{3}=-3+2 \sqrt{2}, \quad x_{4}=-3-2 \sqrt{2}
$$

and the corresponding $y$-values are:

$$
y_{1}=3, \quad y_{2}=-2, \quad y_{3}=-3-2 \sqrt{2}, \quad y_{4}=-3+2 \sqrt{2}
$$

Results from the 1953 Rákosi Mátyás Mathematical Competition, First Round, by counties and school types

| Counties and Budapest | Submitted Papers |  |  | Advanced to Finals |  |  |
| :--- | :---: | :---: | :---: | :---: | :---: | :---: |
|  | High School | Voc. Tech. | Total |  |  |  |
|  | High School | Voc. Tech. |  |  |  |  |
| 1. Baranya | 93 | 39 | 132 | 7 | - | 7 |
| 2. Bács-Kiskún | 91 | 4 | 95 | 4 | - | 4 |
| 3. Békés | 147 | 19 | 166 | 10 | 2 | 12 |
| 4. Borsod | 142 | 42 | 184 | 4 | 4 | 8 |
| 5. Csongrád | 94 | 61 | 155 | 8 | 4 | 12 |
| 6. Fejér | 27 | 13 | 40 | 2 | - | 2 |
| 7. Győr-Sopron | 107 | 19 | 126 | 18 | - | 18 |
| 8. Hajdu-Bihar | 100 | 25 | 125 | 4 | 1 | 5 |
| 9. Heves | 29 | 10 | 39 | 4 | - | 4 |
| 10. Komárom | 100 | 26 | 126 | 9 | - | 9 |
| 11. Nógrád | 25 | 12 | 37 | 3 | - | 3 |
| 12. Pest | 68 | 2 | 70 | 3 | - | 3 |
| 13. Somogy | 23 | 6 | 29 | 4 | - | 4 |
| 14. Szabolcs-Szatmár | 68 | 16 | 84 | 2 | - | 2 |
| 15. Szolnok | 106 | 5 | 111 | 5 | 1 | 6 |
| 16. Tolna | 83 | - | 83 | - | - | - |
| 17. Vas | 118 | 3 | 121 | 2 | - | 2 |
| 18. Veszprém | 100 | 5 | 105 | 9 | - | 9 |
| 19. Zala | 28 | 6 | 34 | 3 | - | 3 |
| 20. Budapest | 1011 | 70 | 1081 | 64 | 12 | 76 |
| Total | 2560 | 383 | 2943 | 165 | 24 | 189 |

II. Solution: The second equation can be written as

$$
x y(x+y)+6=0
$$

and notice that the expressions $u=x+y$ and $v=x y$ are the sum and product of the given terms:

$$
u+v=-5, \quad u v=-6
$$

Thus, $u$ and $v$ are the two roots of the equation

$$
t^{2}+5 t-6=0
$$

which are

$$
u_{1}=\frac{-5+7}{2}=1, \quad v_{1}=\frac{-5-7}{2}=-6
$$

or vice versa

$$
u_{2}=-6, \quad v_{2}=1
$$

Since these represent $x+y$ and $x y$, respectively, $x$ and $y$ can be the two roots of the equations

$$
z^{2}-z-6=0
$$

and

$$
z^{2}+6 z+1=0
$$

From the first equation,

$$
x_{1}=3, \quad y_{1}=-2 ; \quad x_{2}=-2, \quad y_{2}=3
$$

From the second equation,

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

$$
\begin{aligned}
& y_{3}=-3-2 \sqrt{2} \\
& y_{4}=-3+2 \sqrt{2}
\end{aligned}
$$