7.283. $\left\{\begin{array}{l}\log _{x}(3 x+2 y)=2, \\ \log _{y}(2 x+3 y)=2 .\end{array}\right.$
## Solution.

Domain of definition: $\left\{\begin{array}{l}00, \\ 2 x+3 y>0 .\end{array}\right.$

Transform the system taking into account the domain of definition

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ 3 x + 2 y = x ^ { 2 } , } \\
{ 2 x + 3 y = y ^ { 2 } }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
y=\frac{x^{2}-3 x}{2}, \\
2 x+3 y-y^{2}=0
\end{array} \Rightarrow y=\frac{x^{2}-3 x}{2},\right.\right. \\
& 2 x+\frac{3\left(x^{2}-3 x\right)}{2}=\left(\frac{x^{2}-3 x}{2}\right)^{2} \Leftrightarrow x^{3}-6 x^{2}+3 x+10=0 \Leftrightarrow \\
& \Leftrightarrow(x+1)(x-2)(x-5)=0, x_{1}=-1, x_{2}=2, x_{3}=5,
\end{aligned}
$$

then $y_{1}=2, y_{2}=-1, y_{3}=5 .\left\{\begin{array}{l}x_{1}=-1, \\ y_{1}=2\end{array}\right.$ and $\left\{\begin{array}{l}x_{2}=2, \\ y_{2}=-1\end{array}\right.$ do not fit the domain of definition.

Answer: $(5 ; 5)$.