3x - xy = y + 10
⇒x.(3-y)=10+y
Nếu: 3-y=0⇒y=3⇒0x=13 (loaij)
Chia 2 vế cho 3-y$\neq$0
⇒x=$\frac{10+y}{3-y}$
⇒x=$\frac{-(3-y)+13}{3-y}$
⇒x=-1+$\frac{13}{3-y}$
⇒3-y∈Ư(13)={±1;±13}
3-y=1⇒y=2⇒x=12 (tm)
3-y=-1⇒y=4⇒x=-14 (tm)
3-y=13⇒y=-10⇒x=0 (tm)
3-y=-13⇒y=16⇒x=-2
Vậy (x,y)∈{(12;2);(-14;4);(0;-10);(-2;16)}