3B. In the set of integers, solve the equation

$$
\frac{x+6}{y}+\frac{10}{x y}=\frac{y-4}{x}
$$
Solution. Clearly, $x, y \neq 0$. The given equation is successively equivalent to the equation

$$
\begin{aligned}
& x^{2}+6 x+10=y^{2}-4 y \\
& (y-2)^{2}-(x+3)^{2}=5 \\
& (y-x-5)(y+x+1)=5
\end{aligned}
$$

Since $x, y \in \mathbb{Z}$, from the last equation we obtain the systems of equations

$$
\left\{\begin{array} { l } 
{ y - x - 5 = 1 } \\
{ y + x + 1 = 5 }
\end{array} \quad \left\{\begin{array} { l } 
{ y - x - 5 = - 1 } \\
{ y + x + 1 = - 5 }
\end{array} \quad \left\{\begin{array} { l } 
{ y - x - 5 = 5 } \\
{ y + x + 1 = 1 }
\end{array} \quad \left\{\begin{array}{l}
y-x-5=-5 \\
y+x+1=-1
\end{array}\right.\right.\right.\right.
$$

whose solutions are

$$
(x, y)=(-1,5) ;(x, y)=(-5,-1) ;(x, y)=(-5,-5),(x, y)=(-1,-1)
$$

and these are the solutions to the original equation.