Solve the following system of equations:

$$
x^{2}+x y+y^{2}=7
$$

$$
x^{2}+x z+z^{2}=13
$$

$$
y^{2}+y z+z^{2}=19
$$
From (2) subtracting (1) and from (3) subtracting (2), I obtain the following two equations:

$$
(z-y)(x+y+z)=6
$$

$$
(y-x)(x+y+z)=6
$$

From these two equations, it follows that

$$
z-y=y-x
$$

from which

$$
2 y=x+z
$$

Substituting this value into (4), I get:

$$
y(z-y)=2
$$

from which

$$
z=\frac{2+y^{2}}{y}
$$

Substituting this value of $z$ into (3), we get:

$$
3 y^{4}-13 y^{2}-34=0
$$

From this, the values of $y$ are

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

The corresponding values of $z$ and $x$ are obtained from 9) and 7), respectively; they are:

$$
\begin{aligned}
& z_{1}=3, z_{2}=-3, z_{3}=\frac{7}{3} \sqrt{3}, z_{4}=-\frac{7}{3} \sqrt{3} \\
& x_{1}=1, x_{2}=-1, x_{3}=-\frac{5}{3} \sqrt{3}, x_{4}=\frac{5}{3} \sqrt{3}
\end{aligned}
$$

(Friedmann Bernát, secondary school VII. grade, S.-A.-Ujhely.

The problem was also solved by: Geiszt Emil, Schiller Jenő, and Szabó Gusztáv, Győr; Grünhut Béla, Pécs; Kántor Nándor and Pósch Gyula, Budapest; Visnya Aladár, Pécs; Zemplén Győző, Fiume.