xy-3x+y=-20
⇒x(y-3)=-y-20
Nếu y-3=0⇒y=3⇒0x=-23 (loại)
CHia 2 vế cho y-3$\neq$0, ta được:
$x=\frac{-y-20}{y-3}$
⇒$x=\frac{-(y-3)-23}{y-3}$
⇒$x=-1-\frac{23}{y-3}
⇒y-3∈Ư(23)={±1;±23}
y-3=1⇒y=4⇒x=-24 (tm)
y-3=-1⇒y=2⇒x=22 (tm)
y-3=23⇒y=26⇒x=-2 (tm)
y-3=-23⇒y=-20⇒x=0 (tm)
Vậy (x,y)∈{(-24;4);(22;2);(-2;26);(0;-20)}