7.137. $\left\{\begin{array}{l}2^{\frac{x-y}{2}}+2^{\frac{y-x}{x}}=2.5 \\ \lg (2 x-y)+1=\lg (y+2 x)+\lg 6\end{array}\right.$
## Solution.

Domain of definition: $\left\{\begin{array}{l}2 x-y>0, \\ y+2 x>0 .\end{array}\right.$

From the first equation of the system, we get $\left(2^{\frac{x-y}{2}}\right)^{2}-2.5 \cdot 2^{\frac{x-y}{2}}+1=0$. Solving this equation as a quadratic equation in terms of $2^{\frac{x-y}{2}}$, we find $\left(2^{\frac{x-y}{2}}\right)_{1}=2^{-1}$ or $\left(2^{\frac{x-y}{2}}\right)_{2}=2$, from which $(x-y)_{1}=-2$ or $(x-y)_{2}=2$. From the second equation of the system, we get $\lg 10(2 x-y)=\lg 6(y+2 x)$, from which $10(2 x-y)=6(y+2 x), x=2 y$. Thus, the original system is equivalent to the systems of equations:

1) $\left\{\begin{array}{l}x-y=-2 \\ x=2 y\end{array}\right.$
2) $\left\{\begin{array}{l}x-y=2 \\ x=2 y\end{array}\right.$

from which:

1) $\left\{\begin{array}{l}x_{1}=-4 \\ y_{1}=-2\end{array}\right.$
2) $\left\{\begin{array}{l}x_{2}=4 \\ y_{2}=2\end{array}\right.$
$\left\{\begin{array}{l}x_{1}=-4 \\ y_{1}=-2 \text { (does not satisfy the domain of definition). }\end{array}\right.$

Answer: $(4 ; 2)$