Ta có:
\(xy - x + 2y = 3\)
\(xy - x + 2y-3 = 0\)
\(xy - x + 2y-3+1 = 1\)
\(x\left(y-1\right)+2\left(y-1\right)=1\)
\(\left(y-1\right).\left(x+2\right)=1\)
\(\Rightarrow\left[\begin{matrix}y-1=1;-1\\x+2=1;-1\end{matrix}\right.\)
\(\Rightarrow\left[\begin{matrix}y-1=1\Rightarrow y=1+1=2\\x+2=1\Rightarrow x=1-2=-1\end{matrix}\right.\)
\(\Rightarrow\left[\begin{matrix}y-1=-1\Rightarrow y=-1+1=0\\x+2=-1\Rightarrow x=-1-2=-3\end{matrix}\right.\)
Vậy \(y=\left\{2;0\right\},x=\left\{-1;-3\right\}\)
Ta có:
xy-x+2y=3
xy-x+2y-3=0
xy-x+2y-3+1=1
x(y-1)+2(y-1)=1
(y-1).(x+2)=1
\(\Rightarrow\) (y-1) và (x+2) lần lượt là các cặp (1;1),(-1;-1)
Ta có: y-1=1
\(\Rightarrow\) y=2
x+2=1
\(\Rightarrow\) x=-1
Mặt khác, ta có:
y-1=-1
\(\Rightarrow\) y=0
x+2=-1
\(\Rightarrow\) x=-3
