Solve the following system of equations:

$$
\begin{aligned}
& 1-\frac{12}{3 x+y}=\frac{2}{\sqrt{x}} \\
& 1+\frac{12}{3 x+y}=\frac{6}{\sqrt{y}}
\end{aligned}
$$
Solution. Assuming $x, y>0$, by adding the two equations and dividing both sides of the obtained equation by 2, we get:

$$
1=\frac{3}{\sqrt{x}}+\frac{1}{\sqrt{y}}
$$

By subtracting the first equation from the second and dividing both sides by 2:

$$
\frac{12}{3 x+y}=\frac{3}{\sqrt{y}}-\frac{1}{\sqrt{x}}
$$

Multiplying equations (1) and (2):

$$
\frac{12}{3 x+y}=\frac{9}{y}-\frac{1}{x}=\frac{9 x-y}{x y}
$$

After bringing to a common denominator:

$$
12 x y=(9 x-y)(3 x+y)=27 x^{2}+6 x y-y^{2}
$$

Factoring the quadratic equation: $(3 x-y) \cdot(9 x+y)=0$. Since $x, y>0$, only $3 x-y=0$ is possible, so $x=\frac{y}{3}$. Substituting this into (1):

$$
1=\frac{\sqrt{3}}{\sqrt{y}}+\frac{3}{\sqrt{y}}
$$

From this, $\sqrt{y}=3+\sqrt{3} \Rightarrow y=12+6 \sqrt{3}$ and $x=4+2 \sqrt{3}$.

Therefore, the solution to the system of equations is: $x=4+2 \sqrt{3} \approx 7.464, y=12+6 \sqrt{3} \approx 22.392$.