7.132. $\left\{\begin{array}{l}3^{y} \cdot 9^{x}=81 \\ \lg (y+x)^{2}-\lg x=2 \lg 3 .\end{array}\right.$
## Solution.

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

From the first equation of the system $3^{y+2 x}=3^{4}, y+2 x=4, y=4-2 x$.

From the second equation of the system $\lg \frac{(y+x)^{2}}{x}=\lg 9$, hence $\frac{(y+x)^{2}}{x}=9$. Then the original system becomes

$$
\left\{\begin{array}{l}
y=4-2 x \\
(y+x)^{2}=9 x
\end{array} \Rightarrow x^{2}-17 x+16=0\right.
$$

from which $x_{1}=1, x_{2}=16$. Then $y_{1}=2, y_{2}=-28$.

Answer: $(1 ; 2),(16 ;-28)$