7.261. $\left\{\begin{array}{l}\left(0,48^{x^{2}+2}\right)^{2 x-y}=1, \\ \lg (x+y)-1=\lg 6-\lg (x+2 y) .\end{array}\right.$

7.261. $\left\{\begin{array}{l}\left(0.48^{x^{2}+2}\right)^{2 x-y}=1, \\ \log (x+y)-1=\log 6-\log (x+2 y) .\end{array}\right.$
Solution.

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

Rewrite the first equation of the system as

$0.48^{\left(x^{2}+2\right)(2 x-y)}=0.48^{0} \Leftrightarrow\left(x^{2}+2\right)(2 x-y)=0 \Leftrightarrow 2 x-y=0, y=2 x$.

From the second equation of the system we have

$\lg \frac{x+y}{10}=\lg \frac{6}{x+2 y} \Leftrightarrow \frac{x+y}{10}=\frac{6}{x+2 y} \Rightarrow \frac{x+2 x}{10}=\frac{6}{x+4 x}, x^{2}=4$.

Then $x_{1}=-2, x_{2}=2 ; y_{1}=-4, y_{2}=4$.

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

Answer: $(2 ; 4)$.