5. [6] Suppose $x$ and $y$ are positive real numbers such that
$$
x+\frac{1}{y}=y+\frac{2}{x}=3 \text {. }
$$

Compute the maximum possible value of $x y$.
Answer:
$$
3+\sqrt{7}
$$

Solution 1: 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
$$
\left(\frac{C+2}{3}\right)\left(\frac{C+1}{3}\right)=C \Longrightarrow C^{2}-6 C+2=0 .
$$

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