Problem 721

The necessary conditions for the given equation system are $x, y > 0; y+3x \neq 0$. The given equation system is equivalent to the following equation systems
$$\begin{cases} 1 - \frac{12}{y+3x} = \frac{2}{\sqrt{x}} \\ 1 + \frac{12}{y+3x} = \frac{6}{\sqrt{x}} \end{cases} \Rightarrow \begin{cases} \frac{1}{\sqrt{x}} + \frac{1}{\sqrt{y}} = 1 \\ -\frac{1}{\sqrt{x}} + \frac{3}{\sqrt{y}} = \frac{12}{y+3x} \end{cases} \quad (1) \quad (2)$$
Multiply (1) with (2), we get
$$\frac{9}{y} - \frac{1}{x} = \frac{12}{y+3x} \Leftrightarrow y^2 + 6xy - 27x^2 = 0 \Leftrightarrow \begin{cases} y = 3x \\ y = -9x \end{cases}$$
By the condition $x, y > 0$, we have $y = 3x$. By setting $y = 3x$ in (1), we obtain
$$\frac{1}{\sqrt{x}} + \frac{3}{\sqrt{3x}} = 1 \Rightarrow x = 4 + 2\sqrt{3}, y = 12 + 6\sqrt{3}$$