7.128. $\left\{\begin{array}{l}\log _{y} x+\log _{x} y=2, \\ x^{2}-y=20\end{array}\right.$

The system of equations is:
\[
\left\{\begin{array}{l}
\log _{y} x+\log _{x} y=2, \\ 
x^{2}-y=20
\end{array}\right.
\]
Solution.

Domain of definition: $\left\{\begin{array}{l}0<x \neq 1, \\ 0<y \neq 1 .\end{array}\right.$

From the first equation, we have:

$\log _{y} x+\frac{1}{\log _{y} x}-2=0, \quad \log _{y}^{2} x-2 \log _{y} x+1=0, \quad\left(\log _{y} x-1\right)^{2}=0 ;$
from which $\log _{y} x=1, x=y$. From the second equation of the system, we have $y^{2}-y-20=0$, from which $y_{1}=-4, y_{2}=5 ; y_{1}=-4$ does not satisfy the domain of definition. Then $x=y=5$.

Answer: $(5 ; 5)$