xy-y+2x=15
(=)y.(x-1)+2x-2=13
(=)y.(x-1)+2.(x-1)=13
(=)(x-1).(y+2)=13
ta có các th :
TH1:\(\left\{{}\begin{matrix}x-1=1\\y+2=13\end{matrix}\right.\left(=\right)\left\{{}\begin{matrix}x=2\\y=11\end{matrix}\right.\)
TH2:\(\left\{{}\begin{matrix}x-1=13\\y+2=1\end{matrix}\right.\left(=\right)\left\{{}\begin{matrix}x=14\\y=-1\end{matrix}\right.\)
TH3:\(\left\{{}\begin{matrix}x-1=-13\\y+2=-1\end{matrix}\right.\left(=\right)\left\{{}\begin{matrix}x=-12\\y=1\end{matrix}\right.\)
TH4:\(\left\{{}\begin{matrix}x-1=-1\\y+2=-13\end{matrix}\right.\left(=\right)\left\{{}\begin{matrix}x=0\\y=-15\end{matrix}\right.\)
Vậy có 4 cặp (x,y): (2,11),(14,-1),(-12,1),(0,-15)
