\(x-xy+3y=6\\ x\left(1-y\right)+3y-3=6-3\\ \left(-x\right)\left(y-1\right)+3\left(y-1\right)=3\\ \left(3-x\right)\left(y-1\right)=3\)
Ta có bảng sau:
\(3-x\) | 1 | 3 | -1 | -3 |
\(y-1\) | 3 | 1 | -3 | -1 |
x | 2 | 0 | 4 | 6 |
y | 4 | 2 | -2 | 0 |
Vậy \(\left(x;y\right)\in\left\{\left(2;4\right);\left(0;2\right);\left(4;-2\right);\left(6;0\right)\right\}\)