6.094. $\left\{\begin{array}{l}\frac{x+y}{x-y}+\frac{x-y}{x+y}=\frac{13}{6} \\ x y=5 .\end{array}\right.$
## Solution.

Domain of definition: $x \neq \pm y$.

Transforming the first equation of the system, we get

$$
6(x+y)^{2}+6(x-y)^{2}=13(x-y)(x+y) \Leftrightarrow x^{2}=25 y^{2}
$$

from which

$x_{1}=-5 y, x_{2}=5 y$.

From the second equation of the system, we find $y^{2}=-1$ (not suitable) or $y^{2}=1$, from which $y_{1,2}= \pm 1$. Then $x_{1,2}= \pm 5$.

Answer: $(5 ; 1),(-5 ;-1)$.