Question

求解2元2次方程组！！！

求解此方程组：
X^2+Y^2=7225
（140-X）^2+（Y-12.5）^2=9506.25
貌似有4个解求高手帮忙解答多谢，最好能把过程写一下！

Answer 1

X^2+Y^2=7225
(140-X)^2+(Y-12.5)^2=9506.25
相减
280x-140^2+25y-12.5^2= - 2281.25
280x-25y = 17475
56x-5y = 3495
5y = 3495 - 56x
25x^2+(5y)^2 = 7225*25
25x^2+(3495 - 56x )^2 = 7225*25
3161x^2-391440x +12034400=0
Δ=1062320000
x1 = 391440+32593.25 / 2*3161 =67
x2 = 391440-32593.25 / 2*3161 = 56.76
y的4个值代这里面就行了5y = 3495 - 56x

Answer 2

（140-X）^2+（Y-12.5）^2=9506.25 与X^2+Y^2=7225相减
-280x+140^2-25y^2+12.5^2=2281.25
再用x表示y，带入其中一个方程，求出x,再求出y.