8.5. Determine the numbers $x, y \in R$ for which $3 x=2 y+\frac{3}{y}$ and $3 y=2 x+\frac{3}{x}$.
Solution: 8.5.: (1) We observe that $x \neq 0, y \neq 0$. Adding the equations term by term, we get:

$\left(3 x+3 y=2 x+2 y+\frac{3}{x}+\frac{3}{y}\right) \Leftrightarrow\left(x+y=\frac{3}{x}+\frac{3}{y}\right) \Leftrightarrow$

(2) $\Leftrightarrow\left(x+y=\frac{3(x+y)}{x y}\right) \Leftrightarrow\left(\frac{x+y}{3}=\frac{x+y}{x y}\right) \Leftrightarrow\left[\begin{array}{l}x+y=0 \\ x y=3\end{array}\right.$

(3) From $x+y=0$ we get $y=-x$. Substituting this into the equation $3 y=2 x+\frac{3}{x}$ we obtain $\left(-3 x=2 x+\frac{3}{x}\right) \Leftrightarrow\left(-5 x=\frac{3}{x}\right) \Leftrightarrow\left(x^{2}=-\frac{3}{5}\right) \Leftrightarrow x \in \varnothing$

(4) If $x y=3$, then multiplying the equation $3 y=2 x+\frac{3}{x}$ by $x$ we get

$\left(3 x y=2 x^{2}+3\right) \Leftrightarrow\left(9=2 x^{2}+3\right) \Leftrightarrow\left(x^{2}=3\right) \Leftrightarrow(x \in\{-\sqrt{3}, \sqrt{3}\})$. From $x y=3$ we get $y \in\{-\sqrt{3}, \sqrt{3}\}$. Therefore, the solution is: $(x, y) \in\{(-\sqrt{3},-\sqrt{3}),(\sqrt{3}, \sqrt{3})\}$.