\section*{Problem 3 - 151213}

Determine all triples \((x, y, z)\) of real numbers for which

\[
\begin{aligned}
\frac{1}{x+y}+\frac{1}{x+z} & =\frac{7}{12} \\
\frac{1}{x+y}+\frac{1}{y+z} & =\frac{8}{15} \\
\frac{1}{x+z}+\frac{1}{y+z} & =\frac{9}{20} \quad \text{holds.}
\end{aligned}
\]
Assume \((x, y, z)\) is a solution to the system of equations. Then, from (1), (2), (3) by adding and halving, we get

\[
\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{x+z}=\frac{47}{60}
\]

Subtracting (1) or (2) or (3) from (4), we obtain

\[
\frac{1}{y+z}=\frac{1}{5} \quad \text { or } \quad \frac{1}{x+z}=\frac{1}{4} \quad \text { or } \quad \frac{1}{x+y}=\frac{1}{3}
\]
which implies \(x+y=3, y+z=5, x+z=4\). Adding these three equations and dividing by 2, we get

\[
x+y+z=\frac{1}{2}(3+4+5)=6
\]

From this, we can derive, using the same method as described above,

\[
x=(x+y+z)-(y+z)=1 \quad ; \quad y=2 \quad ; \quad z=3
\]

Thus, the triplet \((1,2,3)\) can be the only solution. Indeed,

\[
\frac{1}{3}+\frac{1}{4}=\frac{7}{12} \quad ; \quad \frac{1}{3}+\frac{1}{5}=\frac{8}{15} \quad ; \quad \frac{1}{4}+\frac{1}{5}=\frac{9}{20}
\]

Taken from \([5]\)