Solve the following system of equations:

$$
\begin{aligned}
& \frac{1}{x}+\frac{1}{x+y}=\frac{1}{a}, \\
& \frac{1}{y}+\frac{1}{z+x}=\frac{1}{b} \\
& \frac{1}{z}+\frac{1}{x+y}=\frac{1}{c}
\end{aligned}
$$
Our system of equations only makes sense if $a \neq 0, b \neq 0, c \neq 0$, and since the reciprocal values of these three constants are also not 0, it follows that $x+y+z \neq 0$. Additionally, we must exclude the cases where $x=0, y=0, z=0, y+z=0, z+x=0$, and $x+y=0$.

From the three equations, we can express the constants as follows:

$$
\begin{aligned}
& a=\frac{x y+x z}{x+y+z} \\
& b=\frac{x y+y z}{x+y+z} \\
& c=\frac{x z+y z}{x+y+z}
\end{aligned}
$$

From these, we get:

$$
\begin{aligned}
a+b-c & =\frac{2 x y}{x+y+z} \\
a-b+c & =\frac{2 x z}{x+y+z} \\
-a+b+c & =\frac{2 y z}{x+y+z}
\end{aligned}
$$

The values of the unknowns can be determined, for example, as follows:

Dividing (9) by (8) and (7), respectively:

$$
y=\frac{-a+b+c}{a-b+c} x, \quad z=\frac{-a+b+c}{a+b-c} x
$$

Substituting these values of $y$ and $z$ into (1):

$$
\frac{1}{x}\left(1+\frac{1}{\frac{-a+b+c}{a-b+c}+\frac{-a+b+c}{a+b-c}}\right)=\frac{1}{a}
$$

From which we get:

$$
x=a+\frac{a(a-b+c)(a+b-c)}{(-a+b+c) 2 a}=\frac{2(a b+a c+b c)-\left(a^{2}+b^{2}+c^{2}\right)}{2(-a+b+c)}
$$

Similarly, we find that:

$$
y=\frac{2(a b+a c+b c)-\left(a^{2}+b^{2}+c^{2}\right)}{2(a-b+c)}, \quad z=\frac{2(a b+a c+b c)-\left(a^{2}+b^{2}+c^{2}\right)}{2(a+b-c)}
$$

Béla Szeidl (Bp. V., Cukor u. g. IV. o. t.)