## Task A-1.3.

Determine all triples of real numbers $(x, y, z)$ such that

$$
\frac{1}{x}+\frac{1}{y+z}=\frac{1}{3}, \quad \frac{1}{y}+\frac{1}{z+x}=\frac{1}{5}, \quad \frac{1}{z}+\frac{1}{x+y}=\frac{1}{7}
$$
## Solution.

From the first equation, by reducing to a common denominator, we get $x y + x z = 3(x + y + z)$. Similarly, we write the other two equations, and we obtain the system in a slightly different form:

$$
\begin{aligned}
& x y + x z = 3(x + y + z) \\
& y z + y x = 5(x + y + z) \\
& z x + y z = 7(x + y + z)
\end{aligned}
$$

The system can be solved for $x y, y z, z x$ as follows. If we add (1) and (2), and subtract (3) from the sum, we get $2 x y = x + y + z$. Similarly, we get $2 y z = 9(x + y + z)$, $2 z x = 5(x + y + z)$.

Thus, it holds that $x + y + z = 2 x y = \frac{2}{9} y z = \frac{2}{5} z x$.

Since from the given equations we know $x \neq 0, y \neq 0$, and $z \neq 0$, we conclude $9 x = 5 y = z$.

By substituting into the first equation, we get $x = \frac{59}{18}$. We also have $y = \frac{59}{10}$, $z = \frac{59}{2}$.

By verification, we easily confirm that

$$
(x, y, z) = \left(\frac{59}{18}, \frac{59}{10}, \frac{59}{2}\right)
$$

is indeed a solution to the given system.