7.270. $\left\{\begin{array}{l}\lg \left(x^{2}+y^{2}\right)=1+\lg 8, \\ \lg (x+y)-\lg (x-y)=\lg 3 .\end{array}\right.$
## Solution.

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

From the condition we have $\left\{\begin{array}{l}\lg \left(x^{2}+y^{2}\right)=\lg 80, \\ \lg \frac{x+y}{x-y}=\lg 3\end{array} \Leftrightarrow\left\{\begin{array}{l}x^{2}+y^{2}=80, \\ \frac{x+y}{x-y}=3 .\end{array}\right.\right.$ From the second equation of the system $x=2 y$. Then from the first equation we have $(2 y)^{2}+y^{2}=80, y^{2}=16$. Hence $y_{1}=-4, y_{2}=4$. Then $x_{1}=-8, x_{2}=8$.

$$
\left\{\begin{array}{l}
x_{1}=-8 \\
y_{1}=-4
\end{array}\right. \text { does not satisfy the domain of definition. }
$$

Answer: $(8 ; 4)$.