22. [12] Let $x<y$ be positive real numbers such that
$$
\sqrt{x}+\sqrt{y}=4 \text { and } \sqrt{x+2}+\sqrt{y+2}=5 .
$$

Compute $x$.
Answer:
$\frac{49}{36}$
Solution: Adding and subtracting both equations gives
$$
\begin{array}{l}
\sqrt{x+2}+\sqrt{x}+\sqrt{y+2}+\sqrt{y}=9 \\
\sqrt{x+2}-\sqrt{x}+\sqrt{y+2}-\sqrt{y}=1
\end{array}
$$

Substitute $a=\sqrt{x}+\sqrt{x+2}$ and $b=\sqrt{y}+\sqrt{y+2}$. Then since $(\sqrt{x+2}+\sqrt{x})(\sqrt{x+2}-\sqrt{x})=2$, we have
$$
\begin{array}{c}
a+b=9 \\
\frac{2}{a}+\frac{2}{b}=1
\end{array}
$$

Dividing the first equation by the second one gives
$$
a b=18, a=3, b=6
$$

Lastly, $\sqrt{x}=\frac{\sqrt{x+2}+\sqrt{x}-(\sqrt{x+2}-\sqrt{x})}{2}=\frac{3-\frac{2}{3}}{2}=\frac{7}{6}$, so $x=4 \frac{49}{36}$.