C++ 求解具有两个未知数的两个方程组

Solve system of two equations with two unknowns

提问by user2925251

Solve the system of two equations with two unknowns below:

求解具有以下两个未知数的两个方程组：

a1, b1, c1, a2, b2 and c2 are inputted by the user himself.

a1、b1、c1、a2、b2、c2由用户自己输入。

I've been trying to find a math solution for the problem first and I can't seem to go far..

我一直试图首先为这个问题找到数学解决方案，但我似乎走不远。

What I've tried so far is :

到目前为止我尝试过的是：

1. From first equation to find y. (b1y = c1-a1x, y = (c1-a1x)/b1)
2. Then I replace y in the second equation and I get one equation with 1 unknown in this case x. However, I can't solve the equation, I get some odd numbers / equations and stopped here.

1. 从第一个方程找到y。(b1y = c1-a1x, y = (c1-a1x)/b1)
2. 然后我在第二个方程中替换 y，在这种情况下，我得到一个未知数为 1 的方程。但是，我无法解方程，我得到了一些奇数/方程并停在这里。

Is this correct or is there an easier way to do this?

这是正确的还是有更简单的方法来做到这一点？

Current code:

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;
}

回答by 4pie0

we solve the linear system using Cramer's rule:

我们使用Cramer 规则求解线性系统：

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_equation_system 1 2 5 1 -1 -1

Cramer equations system: result, x = 1.000000, y = 2.000000

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

回答by gregn3

Javascript version, inspired by 4pie0's answer

Javascript 版本，灵感来自4pie0 的答案

// throws error if intersection can't be found
function intersect_2_lines (
    a,b,e,
    c,d,f
)
{
        /* we solve the linear system
         * ax+by=e
         * cx+dy=f
         */

        var determinant = a*d - b*c;
        if(determinant != 0) {
            var x = (e*d - b*f)/determinant;
            var y = (a*f - e*c)/determinant;
            console.log(`Cramer equations system: result, x = ${x}, y = ${y}\n`);
        } else {
            throw new Error("Cramer equations system: determinant is zero\n" +
                    "there are either no solutions or many solutions exist.\n"); 
        }
        return [x,y];
}