Solve the following system of equations:

$$
\frac{x y}{x+y}=\frac{1}{3}, \quad \frac{y z}{y+z}=\frac{1}{4}, \quad \frac{z x}{z+x}=\frac{1}{5} .
$$
I. Solution. It is clear that in (1)-(3), neither $x, y, z$ nor the denominators can be 0. Taking the reciprocal of both sides of the equations and dividing term by term on the left sides, we obtain a system of linear equations in the unknowns $\frac{1}{x}=u, \frac{1}{y}=v, \frac{1}{z}=w$:

$$
\frac{x+y}{x y}=\frac{1}{y}+\frac{1}{x}=v+u=3
$$

and similarly $w+v=4, u+w=5$.

Subtracting the third equation from the sum of the first two, we get $2 v=2$, so $v=1$, and thus from the first two equations, $u=2$, $w=3$, and finally

$$
x=\frac{1}{2}, \quad y=1, \quad z=\frac{1}{3}
$$

Makai Endre (Budapest, Eötvös J. g. I. o. t.)

II. Solution. The solution is not much more difficult even without the above substitution. Express $x$ and $z$ from (1) and (2), respectively:

$$
\begin{aligned}
& x=\frac{y}{3 y-1} \\
& z=\frac{y}{4 y-1}
\end{aligned}
$$

and substitute these into (3). Simplifying the left side by $y \neq 0$ and expanding by $(3 y-1)(4 y-1)$,

$$
\frac{\frac{y}{3 y-1} \cdot \frac{y}{4 y-1}}{\frac{y}{3 y-1}+\frac{y}{4 y-1}}=\frac{\frac{1}{3 y-1} \cdot \frac{y}{4 y-1}}{\frac{1}{3 y-1}+\frac{1}{4 y-1}}=\frac{y}{7 y-2}=\frac{1}{5}
$$

from which $y=1$. Now from (4) and (5), $x=1 / 2, z=1 / 3$, and this result shows that the above expansion was permissible, we did not expand by 0.

Marosi Judit (Budapest, Berzsenyi D. lg. I. o. t.)