7.275.

$$
\left\{\begin{array}{l}
(x+y) \cdot 3^{y-x}=\frac{5}{27} \\
3 \log _{5}(x+y)=x-y
\end{array}\right.
$$
Solution.

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

Taking the logarithm of the first equation to the base 5, we have

$\log _{5}(x+y) \cdot 3^{y-x}=\log _{5} \frac{5}{27} \Leftrightarrow \log _{5}(x+y)-(x-y) \log _{5} 3=1-3 \log _{5} 3$.

We obtain the system

$$
\begin{aligned}
& \left\{\begin{array}{l}
\log _{5}(x+y)-(x-y) \log _{5} 3=1-3 \log _{5} 3, \\
3 \log _{5}(x+y)=x-y
\end{array} \Rightarrow\right. \\
& \Rightarrow \log _{5}(x+y)-3 \log _{5}(x+y) \cdot \log _{5} 3=1-3 \log _{5} 3 \Leftrightarrow \\
& \Leftrightarrow \log _{5}(x+y) \cdot\left(1-3 \log _{5} 3\right)=1-3 \log _{5} 3 \Leftrightarrow \log _{5}(x+y)=1,
\end{aligned}
$$

$$
\text { hence } x+y=5 . \text { Then } x-y=3 \log _{5} 5=3 \text {. From this }\left\{\begin{array}{l}
x+y=5, \\
x-y=3
\end{array},\left\{\begin{array}{l}
x=4 \\
y=1
\end{array}\right.\right.
$$

Answer: $(4 ; 1)$.