a. ta co \(x+y=-2=>x=-2-y\left(1\right)\)
Thay (1) vào xy = -15 ta được \(y.\left(-2-y\right)=-15\)
| \(y\) | - 1 | - 3 | - 5 | - 15 | 1 | 3 | 5 | 15 |
| \(x=\)\(-2-y\) | - 15 | - 5 | - 3 | - 1 | 15 | 5 | 3 | 1 |
| \(x\) | 13 | 3 | 1 | - 1 | - 17 | - 7 | - 5 | - 3 |
b.
| x - 2 | -10 | -5 | -2 | -1 | 1 | 2 | 5 | 10 |
| x | -8 | -3 | 0 | 1 | 3 | 4 | 7 | 12 |
| 2y + 1 | -1 | -2 | -5 | -10 | 10 | 5 | 2 | 1 |
| y | - 1 | \(\dfrac{-3}{2}\) | - 3 | \(\dfrac{-11}{2}\) | \(\dfrac{9}{2}\) | 2 | \(\dfrac{1}{2}\) | 0 |
c.
\(xy-3x+2y=11\)
\(=>x.\left(y-3\right)+2.\left(y-3\right)+6=11\)
\(=>\left(x+2\right).\left(y-3\right)=5\)
| \(x+2\) | - 5 | - 1 | 1 | 5 |
| x | - 7 | - 3 | -1 | 3 |
| \(y-3\) | - 1 | - 5 | 5 | 1 |
| y | 2 | - 2 | 8 | 4 |
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