118. Solve the equation in integers:

a) $x y + 3 x - 5 y = -3$;

b) $x - y = \frac{x}{y}$.
Solution. a) To factor the left side of the equation, subtract 15 from both sides of the equation. We will have:

$$
\begin{gathered}
x y+3 x-5 y-15=-3-15 \\
x(y+3)-5(y+3)=-18 \\
(x-5)(y+3)=-18
\end{gathered}
$$

Now we get the following 12 systems of equations:
$\left\{\begin{array}{l}x-5=-1 \\ y+3=18\end{array}\right.$
$\left\{\begin{array}{l}x-5=-2 \\ y+3=9\end{array}\right.$
$\left\{\begin{array}{l}x-5=-3, \\ y+3=6 ;\end{array}\right.$
$\left\{\begin{array}{l}x-5=1 \\ y+3=-18\end{array}\right.$
$\left\{\begin{array}{l}x-5=2, \\ y+3=-9 ;\end{array}\right.$
$\left\{\begin{array}{l}x-5=3 \\ y+3=-6 ;\end{array}\right.$
$\left\{\begin{array}{l}x-5=-18 \\ y+3=1\end{array}\right.$
$\left\{\begin{array}{l}x-5=-9 \\ y+3=2\end{array}\right.$
$\left\{\begin{array}{l}x-5=-6 \\ y+3=3\end{array}\right.$
$\left\{\begin{array}{l}x-5=18 \\ y+3=-1\end{array}\right.$
$\left\{\begin{array}{l}x-5=9 \\ y+3=-2\end{array}\right.$
$\left\{\begin{array}{l}x-5=6 . \\ y+3=-3 .\end{array}\right.$

Solving each of these systems of equations, we get 12 solutions.

Answer: $(6 ;-21),(-13 ;-2),(4 ; 15),(23 ;-4), \quad(7 ;-12)$, $(-4 ;-1),(3 ; 6),(14 ;-5),(8 ;-9),(-1 ; 0),(2 ; 3),(11 ;-6)$.

b) We have $y(x-y)=x$, or $x y-y^{2}=x$, i.e., $x(y-1)=y^{2}$, from which $x=\frac{y^{2}}{y-1}$.

By the condition, $\frac{y^{2}}{y-1}$ is an integer, so $y-1= \pm 1$. (The numbers $y$ and $y-1$ are coprime, which is easily proven by contradiction: if the numbers $y$ and $y-1$ had a common divisor $d$ different from 1, then $d$ would divide their difference, which is equal to one, which is impossible since $d \neq 1$. Therefore, the fraction $\frac{y^{2}}{y-1}$ is irreducible.)

Thus, $y=2(y \neq 0$ by the condition $), x=4$.

Answer: $(4 ; 2)$.