Solve the following system of equations in the positive real numbers:

$$
x^{2}+y^{2}+x y=7, \quad x^{2}+z^{2}+x z=13, \quad y^{2}+z^{2}+y z=19
$$
I. solution. By subtracting the corresponding sides of the first two equations, it follows that

$$
(z-y)(z+y)+x(z-y)=6, \quad \text { that is } \quad(z-y)(x+y+z)=6
$$

By subtracting the corresponding sides of the second two equations, we get:

$$
(y-x)(y+x)+z(y-x)=6, \quad \text { that is } \quad(y-x)(x+y+z)=6
$$

From (1) and (2), it follows that: $y-x=z-y$, or $z=2 y-x$.

Substituting the latter into the original second equation, and using the first equation, the following system of equations can be written:

$$
\left.\begin{array}{rl}
x^{2}+4 y^{2}-2 x y & =13 \\
x^{2}+y^{2}+x y & =7
\end{array}\right\}
$$

from which we get:

$$
\begin{aligned}
7 x^{2}+28 y^{2}-14 x y & =91 \\
13 x^{2}+13 y^{2}+13 x y & =91
\end{aligned}
$$

Subtracting the upper from the lower and dividing by 3: $2 x^{2}-5 y^{2}+9 x y=0$, which can be factored as: $(x+5 y)(2 x-y)=0$. Since $x+5 y>0$, then $y=2 x, z=2 \cdot 2 x-x=3 x$. Then the original first equation is: $x^{2}+4 x^{2}+2 x^{2}=7$, or $x^{2}=1$.

Since $x>0$, then $x=1, y=2, z=3$. We can verify that these numbers indeed satisfy all three equations.

II. solution. Interpret the equations as the cosine rule written for each of a triangle. In the triangle with sides $\sqrt{7}, x, y$,

$$
(\sqrt{7})^{2}=x^{2}+y^{2}-2 \cos \alpha \cdot x y=x^{2}+y^{2}+x y
$$

Here $-2 \cos \alpha=1$, from which $\cos \alpha=-\frac{1}{2}$, or $\alpha=120^{\circ}$.

Similarly, in the triangles with sides $\sqrt{13}, x, z$ and $\sqrt{19}, y, z$:

$$
\begin{array}{ll}
(\sqrt{13})^{2}=x^{2}+z^{2}-2 \cos \beta \cdot x z, & \text { where } \beta=120^{\circ} \\
(\sqrt{19})^{2}=y^{2}+z^{2}-2 \cos \gamma \cdot y z & \text { where } \gamma=120^{\circ}
\end{array}
$$

In these triangles, therefore, there is a $120^{\circ}$ angle opposite the known sides. By fitting the three triangles together at their $120^{\circ}$ angles, we get a single triangle with sides $\sqrt{7}, \sqrt{13}, \sqrt{19}$. The fitting point will be the isogonal point of the triangle, and a triangle has at most one such point. This means that the equation has at most one solution. However, we can state one solution, which is: $x=1, y=2, z=3$.