Question

如何解xy−2x−2y=20和x^2y^2+ 3x+ 3y=15这个方程组？

Answer 1

xy−2x−2y = 20 (1)
x^2y^2+ 3x+3y = 15 (2)
3(1) + 2(2) , 得 2(xy)^2 + 3xy = 90
(xy-6)(2xy+15) = 0, xy = 6 或 xy = -15/2
xy = 6 代入 (1), 得 x+y = -7, 解得 (-1, -6), (-6, -1)
xy = -15/2 代入 (1), 得 x+y = -55/4,
解得( [√3505-55]/8, -[√3505+55]/8 ), ( -[√3505+55]/8, [√3505-55]/8 )

Answer 2

方程xy-2x-2y=20
xy-2(x+y)=20
x²y²+3(x+y)=15
(x+y)=(xy-20)×1/2
x²y²+3·1/2(xy-20)=15
x²y²+3/2xy-30=15
可以先解出xy是多少