解二元二次方程组

(x+y)(x-2y)=-5 (x-y)(x+2y)=7x+y+xy=7 x^2+y^2+xy=13过程尽请详细~ 万分感谢
2024-11-15 05:58:22
推荐回答(1个)
回答(1):

1.(x+y)(x-2y)=-5 ,即x^2-xy-2y^2=-5 ①
(x-y)(x+2y)=7 , 即x^2+xy-2y^2=7 ②
①+②得 x^2-2y^2=1 ③
②-①得 xy=6 ④
③*6-④得 6x^2-xy-12y^2=0,(2x-3y)(3x+4y)=0
x=3y/2或x=-4y/3
代入④即可求得原方程组的解:
x1=3,y1=2; x2=-3,y2=-2.

2. x+y+xy=7 ①
x^2+y^2+xy=13 ②
①+②得x^2+y^2+2xy+x+y=20,(x+y)^2+(x+y)-20=0
x+y=4或x+y=-5
由x+y=4代入①得xy=3 , 解得x=1,y=3或x=3,y=1;
由x+y=-5代入①得xy=12,此时无实数解.
所以原方程组的解是:
x1=1,y1=3; x2=3,y2=1.