用 C 语言求解一对方程 - 给出奇数结果的程序

Question

First of all, English is not my native language, so I'm not sure if I have the correct mathematical terms, and because of that I was unable to find any previous answers to this question, so apologies in advance if this is a duplicate.

Anyway, I need to write a program in C that is meant to solve a pair of equations in the format of a*x + b*y = c , where x and y have the same value across both equations. The method I'm using consists of dividing all three values (a, b and c) in the first equation by a followed by moving b*y to the other side of the equation so I can get a equation that says x = c - b*y . Afterwards, I insert that value of x into the second equation and from there get the value of y, which I then insert back into the first one and get x.

However, my code is giving strange results - for example, the pair of equations x + 3y = 25 and 2x − 5y = −27 (so, the input would be 1 3 25 2 -5 -27) gives the answers -206 and -77 instead of the correct answers of 4 and 7.

I'm posting the entire code since it's fairly short (<40 lines) and I have no idea where the problem is:

#include<stdio.h>
void lin_jednacina2(float a1, float b1, float c1, float a2, float b2, float c2)
{
    float x, y;

    a1 = 1;
    b1 = b1 / a1;
    c1 /= a1;

    b1 = -1 * b1;

    b2 = b2 - a2 * b1;
    c2 = c2 - a2 * c1;

    y = c2 / b2;
    x = c1 - b1 * y; 

    printf("\nThe values of x and y are %.2f and %.2f, respectively.", x, y);
    return;
}
int main()
{
    float a1, b1, c1, a2, b2, c2;

    printf("Please enter the values of a, b and c for the first equation, in the format of ax + by = c. \n");
    scanf("%f %f %f", &a1, &b1, &c1);
    printf("Please enter the values of a, b and c for the second equation, in the format of ax + by = c. \n");
    scanf("%f %f %f", &a2, &b2, &c2);

    lin_jednacina2(a1, b1, c1, a2, b2, c2);
    return 0;
}

Answer 1

There is something wrong in your's algorithm - it seems like some things are unnecesary divided by a1...

This works ok:

void lin_jednacina2(float a1, float b1, float c1, float a2, float b2, float c2)
{
    float x, y;

    float b11 = b1 / a1;
    float c11 = c1 / a1;
    float a21 = a2 / a1;

    y = (c2 - a21 * c1) / (b2 - a21 * b1); 
    x = c11 - b11 * y;

    printf("\nThe values of x and y are %.2f and %.2f, respectively.", x, y);
    return;
}

Of course, you should add some "fuses" (zero checks) to protect from dividing by 0.