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求解具有两个未知数的两个方程的系统

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求解下面两个未知数的两个方程组的系统：

enter image description here

a1，b1，c1，a2，b2和c2由用户自己输入 .

我一直试图找到问题的数学解决方案，我似乎无法走远...

到目前为止我尝试过的是：

  • 从第一个方程找到y . （b1y = c1-a1x，y =（c1-a1x）/ b1）

  • 然后我在第二个等式中替换y，得到一个方程式，其中1个未知数，在这种情况下为x . 但是，我无法解决这个等式，我得到了一些奇数/方程并停在这里 .

这是正确的还是有更简单的方法来做到这一点？

当前代码：

#include <iostream>

using namespace std;

int main()
{
    int a1, b1, c1, a2, b2, c2;
    cout << "Enter the values for the first equation." << endl;
    cout << "Enter the value for a1" << endl;
    cin >> a1;
    cout << "Enter the value for b1" << endl;
    cin >> b1;
    cout << "Enter the value for c1" << endl;
    cin >> c1;
    cout << "Enter the values for the second equation." << endl;
    cout << "Enter the value for a2" << endl;
    cin >> a2;
    cout << "Enter the value for b2" << endl;
    cin >> b2;
    cout << "Enter the value for c2" << endl;
    cin >> c2;
    cout << "Your system of equations is the following:" << endl;
    cout << a1 << "x+" << b1 << "y=" << c1 << endl;
    cout << a2 << "x+" << b2 << "y=" << c2 << endl;

if ((a1 * b2) - (b1 * a2) == 0){
    cout << "The system has no solution." << endl;
}
else{
    res_x = ((c1*b2) - (b1*c2))/((a1*b2)-(b1*a2));
    res_y = ((a1*c2) - (c1*a2)) / ((a1*b2) - (b1*a2));
    cout << "x=" << res_x << " y=" << res_y << endl;
}

    return 0;
}
c++ math equation equation-solving

1 回答

  • 11

    我们使用Cramer's rule求解线性系统：

    int main(int argc, char** argv) {
    /* we solve the linear system
     * ax+by=e
     * cx+dy=f
     */
    if(argc != 7) {
        cerr<<"Cramer equations system: error,"
                             " we need a,b,c,d,e,f parameters.\n";
        return -1;
    }

    double a,b,e;
    double c,d,f;
    sscanf(argv[1],"%lf",&a);
    sscanf(argv[2],"%lf",&b);
    sscanf(argv[3],"%lf",&e);
    sscanf(argv[4],"%lf",&c);
    sscanf(argv[5],"%lf",&d);
    sscanf(argv[6],"%lf",&f);

    double determinant = a*d - b*c;
    if(determinant != 0) {
        double x = (e*d - b*f)/determinant;
        double y = (a*f - e*c)/determinant;
        printf("Cramer equations system: result, x = %f, y = %f\n", x, y);
    } else {
        printf("Cramer equations system: determinant is zero\n"
                "there are either no solutions or many solutions exist.\n"); 
    }
    return 0;
}

    ./cramer_equation_system 1 2 5 1 -1 -1

    Cramer方程组：结果，x = 1.000000，y = 2.000000

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