Let $x<y$ be positive real numbers such that \[ \sqrt{x}+\sqrt{y}=4 \quad \text { and } \quad \sqrt{x+2}+\sqrt{y+2}=5 \] Compute $x$.
Solution: Adding and subtracting both equations gives \[ \begin{aligned} & \sqrt{x+2}+\sqrt{x}+\sqrt{y+2}+\sqrt{y}=9 \\ & \sqrt{x+2}-\sqrt{x}+\sqrt{y+2}-\sqrt{y}=1 \end{aligned} \] 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 \[ \begin{gathered} a+b=9 \\ \frac{2}{a}+\frac{2}{b}=1 \end{gathered} \] 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 $x=\frac{49}{36}$. $\fbox{\frac{49}{36}}$.