7.269. $\left\{\begin{array}{l}3^{-x} \cdot 2^{y}=1152 \\ \log _{\sqrt{5}}(x+y)=2 .\end{array}\right.$
## Solution.

Domain of definition: $x+y>0$.

Let's write the system of equations as

$$
\left\{\begin{array} { l } 
{ 3 ^ { - x } \cdot 2 ^ { y } = 1152 } \\
{ x + y = 5 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
3^{-x} \cdot 2^{y}=1152 \\
x=5-y
\end{array}\right.\right.
$$

Since $x=5-y$, then

$$
3^{y-5} \cdot 2^{y}=1152 \Leftrightarrow \frac{3^{y} \cdot 2^{y}}{3^{5}}=1152 \Leftrightarrow 6^{y}=279936 \Leftrightarrow 6^{y}=6^{7} \Leftrightarrow y=7
$$

Then $x=-2$.

Answer: $(-2 ; 7)$.