$\left\{ \begin{array}{l}(x+1).(y-1)=2\\(x-3).(y+1)=-6\end{array} \right.$
$⇔\left\{ \begin{array}{l}xy-x+y=3\\xy+x-3y=-3\end{array} \right.$
$⇔2y-x=3$
$⇔x=2y-3$
Khi đó : $(x+1).(y-1) = 2$
$⇔(2y-2).(y-1)=2$
$⇔2.(y-1).(y-1)=2$
$⇔(y-1)^2=1$
$⇔ \left[ \begin{array}{l}y-1=1\\y-1=-1\end{array} \right.$
$⇔ \left[ \begin{array}{l}y=2\\y=0\end{array} \right.$
+) Với $y=0 ⇒(x+1).(-1) = 2$
$⇔x+1=-2$
$⇔x=2$
Với $y=2 ⇒(x+1).1=2$
$⇔x+1=2$
$⇔x=1$
Vậy $(x,y) ∈ \{(2,0);(1,2)\}$
