7.148. $\left\{\begin{array}{l}(x+y) \cdot 2^{y-2 x}=6.25, \\ (x+y) \frac{1}{2 x-y}=5 .\end{array}\right.$
## Solution.

Domain of definition: $\left\{\begin{array}{l}0<x+y \neq 1, \\ 2 x-y \neq 0 .\end{array}\right.$

Taking the logarithm of both equations with base 10, we have

$$
\left\{\begin{array} { l } 
{ \operatorname { l g } ( x + y ) \cdot 2 ^ { y - 2 x } = \operatorname { l g } ( \frac { 5 } { 2 } ) ^ { 2 } , } \\
{ \operatorname { l g } ( x + y ) \frac { 1 } { 2 x - y } = \operatorname { l g } 5 }
\end{array} \Rightarrow \left\{\begin{array}{l}
\lg (x+y)+(y-2 x) \lg 2=2(\lg 5-\lg 2) \\
\frac{\lg (x+y)}{2 x-y}=\lg 5
\end{array}\right.\right.
$$

From the second equation of the system, we get $\lg (x+y)=(2 x-y) \lg 5$, then

$$
\begin{aligned}
& (2 x-y) \lg 5+(y-2 x) \lg 2=2(\lg 5-\lg 2) \quad(2 x-y)(\lg 5-\lg 2)=2(\lg 5-\lg 2) \\
& 2 x-y=2
\end{aligned}
$$

The original system becomes

$$
\left\{\begin{array} { l } 
{ 2 x - y = 2 , } \\
{ \operatorname { l g } ( x + y ) = 2 \operatorname { l g } 5 , }
\end{array} \quad \left\{\begin{array}{l}
2 x-y=2 \\
x+y=25
\end{array}\right.\right.
$$

from which $\left\{\begin{array}{l}x=9 \\ y=16 .\end{array}\right.$

Answer: $(9 ; 16)$