Ceci est trop facile

From $\sqrt{x}+\sqrt{y} = 5\quad \Rightarrow \sqrt{y} = 5 - \sqrt{x}$ .

$\Rightarrow 6 \left( \dfrac{1}{\sqrt{x}} + \dfrac{1}{\sqrt{y}} \right) = 5 \quad \Rightarrow 6 \left( \dfrac{\sqrt{x} + \sqrt{y}} {\sqrt{x} \sqrt{y}} \right) = 5 \quad \Rightarrow 6 \left( \dfrac{5} {\sqrt{x} \sqrt{y}} \right) = 5 \\ \quad \Rightarrow \sqrt{x} \sqrt{y} = 6$

This means that $\sqrt{x}$ and $\sqrt{y}$ are roots of:

$z^2 - 5z+6 = 0 \quad \Rightarrow (z-2)(z-3) = 0 \quad \Rightarrow \begin{cases} z = 2 & \Rightarrow x = 4 \\ z = 3 & \Rightarrow x = 9 \end{cases}$

Therefore, the sum of all possible values of $x$ is $= 4 + 9 = \boxed{13}$ .

By substituting $\sqrt{y} = \sqrt{x} - 5$ in the second equation and then simplifying, you eventually get $X^2 - 5X + 6 = 0$ (where $X = \sqrt{x}$ ). Then it's trivial to solve for $X$ and thereby arrive at the answer.

By inspection, two solution sets Sqrt(x)=2 and Sqrt(y)=3 as well as Sqrt(x)=3 and Sqrt(y)=2 satisfy both equations, hence x=2^2 and 3^2 or 4 and 9, sum is 13.