7.129. $\left\{\begin{array}{l}10^{1+\lg (x+y)}=50, \\ \lg (x-y)+\lg (x+y)=2-\lg 5 .\end{array}\right.$
Solution.

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

## We have:

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ \operatorname { l g } 1 0 ^ { 1 + \operatorname { l g } ( x + y ) } = \operatorname { l g } 5 0 , } \\
{ \operatorname { l g } ( x ^ { 2 } - y ^ { 2 } ) = \operatorname { l g } 2 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
1+\lg (x+y)=\lg 50, \\
x^{2}-y^{2}=20
\end{array} \Leftrightarrow\right.\right. \\
& \Leftrightarrow\left\{\begin{array} { l } 
{ x + y = 5 , } \\
{ ( x - y ) ( x + y ) = 2 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x+y=5, \\
x-y=4,
\end{array}\right.\right.
\end{aligned}
$$

from which $x=\frac{9}{2}, y=\frac{1}{2}$.

Answer: $\left(\frac{9}{2} ; \frac{1}{2}\right)$