7.260. $\left\{\begin{array}{l}3 \cdot\left(\frac{2}{3}\right)^{2 x-y}+7 \cdot\left(\frac{2}{3}\right)^{\frac{2 x-y}{2}}-6=0, \\ \lg (3 x-y)+\lg (y+x)-4 \lg 2=0 .\end{array}\right.$
Solution.

Domain of Definition (DOD): $\left\{\begin{array}{l}3 x-y>0, \\ y+x>0\end{array}\right.$

Solve the first equation of the system as a quadratic equation with respect to $\left(\frac{2}{3}\right)^{\frac{2 x-y}{2}}$. We have

$\left(\frac{2}{3}\right)^{\frac{2 x-y}{2}}=-3$ (no solutions),

$\left(\frac{2}{3}\right)^{\frac{2 x-y}{2}}=\frac{2}{3} \Leftrightarrow \frac{2 x-y}{2}=1, y=2 x-2$.

From the second equation of the system, we have $\lg (3 x-y)(y+x)=\lg 16$, $(3 x-y)(y+x)=16$. We obtain

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ y = 2 x - 2 , } \\
{ ( 3 x - y ) ( y + x ) = 1 6 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
y=2 x-2, \\
(3 x-(2 x-2))(2 x-2+x)=16
\end{array} \Rightarrow\right.\right. \\
& \Rightarrow 3 x^{2}+4 x-20=0 \Rightarrow x_{1}=-\frac{10}{3}, x_{2}=2 .
\end{aligned}
$$

Then $y_{1}=-\frac{26}{3}, y_{2}=2$.

$$
\left\{\begin{array}{l}
x_{1}=-\frac{10}{3} \\
y_{1}=-\frac{26}{3}
\end{array}\right. \text { does not satisfy the DOD. }
$$

Answer: $\quad(2 ; 2)$.