6.242. $\left\{\begin{array}{l}\sqrt{\frac{3 x-2 y}{2 x}}+\sqrt{\frac{2 x}{3 x-2 y}}=2, \\ x^{2}-18=2 y(4 y-9) .\end{array}\right.$
## Solution.

Domain of definition: $\frac{3 x-2 y}{2 x}>0$.

Transform the first equation of the system

$$
\sqrt{\frac{3 x-2 y}{2 x}}+\frac{1}{\sqrt{\frac{3 x-2 y}{2 x}}}-2=0 \Leftrightarrow\left(\sqrt{\frac{3 x-2 y}{2 x}}\right)^{2}-2 \sqrt{\frac{3 x-2 y}{2 x}}+1=0
$$

where $\sqrt{\frac{3 x-2 y}{2 x}} \neq 0,\left(\sqrt{\frac{3 x-2 y}{2 x}}-1\right)^{2}=0 \Rightarrow \sqrt{\frac{3 x-2 y}{2 x}}=1, \frac{3 x-2 y}{2 x}=1, x=2 y$.

From the second equation of the system we have $(2 y)^{2}-18=2 y(4 y-9) \Leftrightarrow$ $\Leftrightarrow 2 y^{2}-9 y+9=0$, from which $y_{1}=\frac{3}{2}, y_{2}=3 ; x_{1}=3, x_{2}=6$.

Answer: $\left(3 ; \frac{3}{2}\right),(6 ; 3)$.