, 2022

Solution:
Rewrite the equations as $x y + 1 = 3 y$ and $x y + 2 = 3 x$. Let $x y = C$, so $x = \frac{C+2}{3}$ and $y = \frac{C+1}{3}$. Then
$$\left(\frac{C+2}{3}\right)\left(\frac{C+1}{3}\right) = C \Longrightarrow C^{2} - 6C + 2 = 0.$$
The larger of its two roots is $3 + \sqrt{7}$.

Solution:
Multiply the two equations to get $x y + 3 + \frac{2}{x y} = 9$, so letting $C = x y$ gives $C^{2} - 6C + 2 = 0$, which has larger root $C = 3 + \sqrt{7}$.