a) 5xy + 5x + 3y = -16
=> 5xy + 5x + 3y + 3 - 3 = -16
=> 5x(y + 1 ) + 3 ( y + 1 ) - 3 = -16
=> ( 5x + 3 ) ( y + 1 ) - 3 = - 16
=> ( 5x + 3 ) ( y + 1 ) = -13
Ta có bảng :
5x + 3 | -13 | 1 | -1 | 13 |
y + 1 | 1 | -13 | 13 | -1 |
=>
x | 2 | |||
y | 0 | -14 | 12 | -2 |
Do x, y E Z => x = 2; y = -2
b) 3x + 7 = y( x + 2)
=> 3x + 7 = xy + 2y
=> 3x + 7 - xy - 2y = 0
=> 3x - xy + 1 + 6 - 2y = 0
=> x ( 3 - y ) + 1 + 2 ( 3 - y ) = 0
=> ( x + 2 ) ( 3 - y ) = -1
Ta có bảng :
x + 2 | 1 | -1 |
3 - y | -1 | 1 |
=>
x | -1 | -3 |
y | 4 | 2 |
Vậy, x = -1; y = 4
hoặc x = -3 ; y =2