求解联立方程的算法 [英] Algorithm for solving simultaneous equations
本文介绍了求解联立方程的算法的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我正在制作一个mfc应用程序,我需要推断两行是否相交。为此,我有2个方程:
x = [-x1y2 + x2y1 - (x2-x1)y] / y1- y2
y = [-x3y4 + x4y3 - (y3-y4)x] / x4-x3
但是我需要一种方法来同时解决这两个方程,我该怎么做?
解决方案
确定,假设 x1,x2,x3,x4,y1,y2,y3,y4
在这个过程中是常量,我们也可以把它写成
x = ab * y
y = cd * x
与 a =( - x1y2 + x2y1)/ y1-y2
等。
现在将第一行代入第二行给出
y = cd *(ab * y)
y(1+ d * b)= cd * a
y =(cd * a)/(1 + d * b)
重新分配给 x = ab * y
给出结果的x部分
I'm making an mfc application in which I need to deduce if two lines intersect or not. For that I have the 2 equations:
x= [-x1y2 +x2y1 - (x2-x1)y ] / y1-y2
y= [-x3y4 +x4y3 - (y3-y4)x ] / x4-x3
But I need a way to solve these 2 equations simultaneously, How would I do that?
解决方案
OK, assuming that x1,x2,x3,x4,y1,y2,y3,y4
are constant inside the process we can also write this as
x=a-b*y
y=c-d*x
with a=(-x1y2+x2y1)/y1-y2
etc.
Now substituting the first line into the second gives
y=c-d*(a-b*y)
y(1+d*b)=c-d*a
y=(c-d*a)/(1+d*b)
resubstituting into x=a-b*y
gives the x part of the result
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