How to solve $xy+ax+by+c=0$ in inetegrs?

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Respected all.

Before I ask your support, let me show you what I have done and have got stuck.

We are willing to solve $2x+3xy+4y=5$. So this is what I have done. The given equation becomes $xy+\frac23x+\frac43=\frac53$.

Let the given equation be $(x+\alpha)(y+\beta)=c$. Then equating we get $\alpha=\frac43, \beta=\frac23, c-\alpha\beta=\frac53$ i.e. $c=-\frac79$. Then we get $(x+\frac43)(y+\frac23)=-\frac79$.

Assume that $x+\frac43=t$ then $y=-\frac23-\frac{7}{9t}$ where $t\in \mathbb Z\backslash \{0\}$. Hence the geenal solution is $$\{(t-\frac43, -\frac23-\frac{7}{9t}: t \in \mathbb Z, t\neq 0)\}$$

Although I do not understand what is the mistake cause when I used MAPLE it said the solutions are ${(-9,-1),(-1,7)}$.

What I am missing here ? Please help me.

P.S. I have got it.

Once we get $(x+\frac43)(y+\frac23)=-\frac{7}{9}$ we multiply both sides by 9 and get $(3x+4)(3y+2)=-7$. Divisors of $-7$ are $\pm1, \pm7$.

Let $3x+4=\pm1$ then we have to consider $3y+2=\mp7$ in order to get $-7$.

Solving we get $(x,y)=(-1,7), (-9,-1)$.