Solve the following system of equations:

$$
\begin{aligned}
& x^{2}-4 \sqrt{3 x-2}+10=2 y \\
& y^{2}-6 \sqrt{4 y-3}+11=x
\end{aligned}
$$
Solution. Let's add the two equations and then rearrange to 0:

$$
x^{2}-4 \sqrt{3 x-2}+10-2 y+y^{2}-6 \sqrt{4 y-3}+11-x=0
$$

Let's try to transform the left-hand side expression into a sum of squares:

$$
((3 x-2)-4 \sqrt{3 x-2}+4)+((4 y-3)-6 \sqrt{4 y-3}+9)+x^{2}-4 x+y^{2}-6 y+13=0
$$

Thus, the equation can be written as:

$$
(\sqrt{3 x-2}-2)^{2}+(\sqrt{4 y-3}-3)^{2}+(x-2)^{2}+(y-3)^{2}=0
$$

A square of a real number is non-negative, so the equality can only hold if \( x=2 \) and \( y=3 \). These values also satisfy the original equations, because

$$
2^{2}-4 \sqrt{3 \cdot 2-2}+10=2 \cdot 3 \quad \text { and } \quad 3^{2}-6 \sqrt{4 \cdot 3-3}+11=2
$$

The solution to the system of equations is: \( x=2 \) and \( y=3 \).