7.130. $\left\{\begin{array}{l}\lg \left(x^{2}+y^{2}\right)=2-\lg 5, \\ \lg (x+y)+\lg (x-y)=\lg 1.2+1 .\end{array}\right.$
Solution.

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

## From the condition

$$
\left\{\begin{array} { l } 
{ \operatorname { l g } ( x ^ { 2 } + y ^ { 2 } ) = \operatorname { l g } 2 0 , } \\
{ \operatorname { l g } ( x ^ { 2 } - y ^ { 2 } ) = \operatorname { l g } 1 2 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x^{2}+y^{2}=20 \\
x^{2}-y^{2}=12
\end{array}\right.\right.
$$

From here, $x^{2}=16$, hence $x_{1,2}= \pm 4 \cdot y^{2}=4, y_{1,2}= \pm 2$. Therefore, $\left\{\begin{array}{l}x_{1}=4, \\ y_{1}=2 ;\end{array}\left\{\begin{array}{l}x_{2}=4 \\ y_{2}=-2 .\end{array}\right.\right.$

The other solutions do not satisfy the DOD.

Answer: $(4 ; 2)(4 ;-2)$