Solve the following system of equations:

$$
x y=x+2 y, \quad y z=y+3 z, \quad z x=z+4 x .
$$
I. solution. In each equation, only two of the three unknowns appear, and in each equation, a different pair. Express $x$ in terms of $y$ from the first equation, and $z$ in terms of $y$ from the second equation, and substitute these into the third equation to get a single-variable equation for $y$:

$$
x=\frac{2 y}{y-1}, \quad z=\frac{y}{y-3}, \quad \frac{2 y^{2}}{(y-1)(y-3)}=\frac{y}{y-3}+\frac{8 y}{y-1} .
$$

By eliminating the fractions and rearranging (assuming of course that $y \neq 1, y \neq 3$),

$$
7 y^{2}-25 y=y(7 y-25)=0
$$

from which $y_{1}=0, y_{2}=25 / 7$ (not excluded values), and consequently the system has two solutions:

$$
\begin{array}{lll}
x_{1}=0, & y_{1}=0, & z_{1}=0 \\
x_{2}=\frac{25}{9}, & y_{2}=\frac{25}{7}, & z_{2}=\frac{25}{4}
\end{array}
$$

Both value sets are indeed solutions.

Rásó István (Ajka, Bródy I. g. II. o. t.)

II. solution. In one of the equations, there is no constant term (furthermore, division by an unknown does not occur), so $x_{1}=0, y_{1}=0, z_{1}=0$ is a solution to the system of equations. We seek a solution different from this. This can only be a solution where none of the unknowns is 0, because if, for example, $x=0$, then from the first and third equations, 0 is obtained for $y$ and $z$ as well. We can reason similarly if $y=0$, or $z=0$.

Divide the first equation by $x y$, the second by $y z$, and the third by $z x$, thus obtaining a system of linear equations in the new unknowns

$$
\frac{1}{x}=t, \quad \frac{1}{y}=u, \quad \frac{1}{z}=v
$$

$$
1=u+2 t, \quad 1=v+3 u, \quad 1=t+4 v
$$

From the first, $u=1-2 t$, substituting this into the second, and expressing $v$ and substituting into the third,

$$
1=v+3-6 t, \quad v=6 t-2, \quad 9=25 t
$$

from which

$$
t=\frac{9}{25}, \quad v=\frac{4}{25}, \quad u=\frac{7}{25}
$$

from which we obtain the above $x_{2}, z_{2}, y_{2}$ solution.