Solve the following system of equations:

$$
\begin{aligned}
& (x+2 y)(x+3 y)=x+y \\
& (2 x+y)(3 x+y)=-99(x+y)
\end{aligned}
$$
I. Solution: Notice that in one of the equations, there is no term that does not contain the unknowns. This implies that the value pair $x=y=0$ is a solution to the system. However, there is no solution where only one of the unknowns is $0$; for instance, the assumption $x=0, y \neq 0$ leads to the different values $y=1/6$ and $y=-99$ from the two equations, and similarly, the assumption $x \neq 0, y=0$ leads to $x=1$ and $x=16.5$. Therefore, we can assume in the following that $x \neq 0$, $y \neq 0$.

Instead of the usual elimination, we eliminate the linear terms from the system; for this, after multiplying the left sides:

$$
\begin{aligned}
& x^{2}+5 x y+6 y^{2}=x+y \\
& 6 x^{2}+5 x y+y^{2}=-99(x+y)
\end{aligned}
$$

we add 99 times the first equation to the second:

$$
105 x^{2}+500 x y+595 y^{2}=0
$$

From this homogeneous equation, by dividing by $5 y^{2}$, we get an equation for the ratio of the two unknowns $x / y=z$:

$$
21 z^{2}+100 z+119=0, \quad \text { from which } \quad z=\left\{\begin{array}{r}
-7 / 3 \\
-17 / 7
\end{array}\right.
$$

Now, with $x=z y=-7 y / 3$, from any of the equations we get $y=6$, and thus $x=-14$, and similarly with $x=-17 y / 7$, $y=35 / 6$, and thus $x=-85 / 6$.

Therefore, the equation has 3 solutions:

$$
\begin{array}{lll}
x_{1}=0, & x_{2}=-14, & x_{3}=-85 / 6 \\
y_{1}=0, & y_{2}=6, & y_{3}=35 / 6
\end{array}
$$

Zalán Péter (Aszód, Petőfi S. g. II. o. t.)

II. Solution: We use the fact that in the second equation under (1), the coefficient of the $x y$ term is the same as in the first, while the coefficients of $x^{2}$ and $y^{2}$ are swapped. Thus, by subtraction, the $x y$ terms cancel out, and the left side can be factored:

$$
-5 x^{2}+5 y^{2}=5(y-x)(x+y)=100(x+y)
$$

from which, by simplification and reduction to zero,

$$
(x+y)(y-x-20)=0
$$

and thus either

$$
\begin{gathered}
x+y=0, \quad \text { or } \\
y-x-20=0
\end{gathered}
$$

From (2), $y=-x$ leads to $x_{1}=0$, and thus $y_{1}=0$.

$$
\text { From (3), } y=x+20, \quad \text { so } \quad 6 x^{2}+169 x+1190=0
$$

and $x_{2}=-14, x_{3}=-85 / 6$, which again leads to the same results.

Pellionisz András (Budapest, Apáczai Csere J. gyak. g. II. o. t.)