$xy +2x-3y=9$
⇔$x$($y+2$)- $3y=9$
⇔$x$($y+2$)- $3y- 3.2=9$
⇔$x$($y+2$)- $3$($y+2$)$=3$
⇔($x-3$)($y+2$) = $3$
⇒$x-3$ và $y+2$ ∈ Ư($3$)={$±1;±3$)
Ta có bảng:
$x-3$ $-3$ $-1$ $1$ $3$
$y+2$ $-1$ $-3$ $3$ $1$
$x$ $0$ $2$ $4$ $6$
$y$ $-3$ $-5$ $1$ $-1$
Vậy ($x;y$)=($0;-3$);($2;-5$);($4;1$);($6;-1$)
