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解二元二次方程组 x^2-xy=6,xy+y^2=4
此题原题是英文:
Find all ordered pairs (x,y) that satisfy both x^2-xy=6 and xy+y^2=4.
人气:409 ℃ 时间:2019-08-18 10:36:03
解答
消去常数项:
1式*2-2式*3:2x²-2xy-3xy-3y²=0
即2x²-5xy-3y²=0
(2x+y)(x-3y)=0
得x=-y/2,或x=3y
将x=-y/2代入方程2:-y²/2+y²=4,得:y²=8,得y=2√2,-2√2
将x=3y代入方程2:3y²+y²=4,得y²=1,得:y=1,-1
因此共有4组(-√2,2√2),(√2,-2√2),(3,1),(-3,-1)
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