C language solution of 3-yuan 1-time equations using the most basic joint elimination method of _c language in primary and secondary schools

I've never learned linear algebra, but many of these algorithms are related to matrices, so we just have to learn to bite the bullet.
Recently I think I can write a linear equation set of procedures? Then thought of such a method, temporarily can only calculate 3 yuan, arbitrary element of the next continue to think. There are too many hard coded, I hope interested readers can give some suggestions to modify!

Copy Code code as follows:

#include "stdafx.h"//vs2010 need
#include "stdio.h"
#include "Stdlib.h"
#include "math.h"
Double x[3];//storage solution x,y,z, using arrays for easy output
X=1,y=2,z=3
Double a[3][4]=
//{
1,1,1,-6,
1,1,-1,0,
1,-1,1,-2
//};
x=2,y=5,z=10
Double a[3][4]=
{
1,2,3,-42,
2,-1,5,-49,
-1,3,-3,17
};

The first equation of a is x,y with 2 and 3 equations, and then the two equations of the equation are obtained and deposited into B
Double b[2][4];
The two equations of B are combined to eliminate Y and get X
Double c[1][4];

void Combineb (int i)//a[0] and a[1] or a[2] elimination Z
{

Double M0=abs (a[i][2]);
int b=a[0][2]*a[i][2]>0?-1:1;
Double Mi=b*abs (a[0][2]);
printf ("m0=%lf,mi=%lf\n", m0,mi);//Trial
for (int j=0;j<4;j++)
{
B[i-1][j]=a[0][j]*m0+a[i][j]*mi;
}

}

void Combinec ()//b[0] and b[1] elimination y
{

Double M0=abs (b[1][1]);
int b=b[0][1]*b[1][1]>0?-1:1;
Double M1=b*abs (b[0][1]);

for (int j=0;j<4;j++)
{
C[0][J]=B[0][J]*M0+B[1][J]*M1;
}

}
Display 4 parameters
void Show (double n[][4],int D1)
{
Char w[3]={' x ', ' y ', ' z '};

for (int i=0;i<d1;i++)
{
for (int j=0;j<3;j++)
{
printf ("%lf *%c +", n[i][j],w[j]);
}
printf ("%LF = 0\n", n[i][3]);
}
printf ("\ n \ nthe");
}

int _tmain (int argc, _tchar* argv[])
{
Show (a,3);
Combineb (1);
Combineb (2);
Show (b,2);
Combinec ();
Show (c,1);

Get x and then count the other numbers
X[0]= (0-c[0][3])/c[0][0];
X[1]= (0-b[0][3]-b[0][0]*x[0])/b[0][1];
X[2]= (0-a[0][3]-a[0][0]*x[0]-a[0][1]*x[1])/a[0][2];
Output results
for (int i=0;i<3;i++)
printf ("x[%d]=%lf\t", I,x[i]);
printf ("\ n");
System ("pause");
return 0;
}

2. Screenshot