2B. Determine all solutions ( $x, y$ ) of the equation $\frac{x+6}{y}+\frac{13}{x y}=\frac{4-y}{x}$, in the set of real numbers.
Solution. It is clear that $x \neq 0$ and $y \neq 0$. Now, if we multiply the equation by $xy$, it takes the form $x^{2}+6 x+13=4 y-y^{2}$, which is equivalent to the equation $x^{2}+6 x+y^{2}-4 y+13=0$.

The left side of the last equation can be written as the sum of two squares: $(x+3)^{2}+(y-2)^{2}=0$. The sum of the squares of two numbers is zero if and only if each of the numbers is zero. Therefore, $x+3=0$ and $y-2=0$, i.e., $x=-3$ and $y=2$.

Thus, the equation has a unique solution $(x, y)=(-3,2)$.