求解兩個方程的系統具有兩個未知數

enter image description here

A1，B1，C1，A2，B2和C2由用戶自己輸入。

我一直試圖先找到該問題的數學解決方案，我似乎無法走多遠..

我試過到目前爲止是：

從第一等式來找到y。（b1y = c1-a1x，y =（c1-a1x）/ b1）
然後我用第二個方程代替y，得到一個方程，其中x爲1。但是，我無法解決方程式，我得到一些奇數/方程式並停在這裏。

這是正確的還是有更簡單的方法來做到這一點？

當前代碼：

#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; 
}

來源

2013-10-27 user2925251

'C++'代碼請 – P0W

在您的代碼，首先，你應該檢查你的2個未知數的系統是否有一個無窮大或沒有解（計算行列式） – lolando

該解直接作爲2×2矩陣的逆矩陣給出（a1，b1; a2，b2），如果矩陣是可逆的（即det！= 0）。 –

我們解決使用Cramer's rule線性系統：

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

克拉默方程系統：因此， x = 1.000000，y = 2.000000

來源

2013-10-27 15:37:58 4pie0

求解兩個方程的系統具有兩個未知數

回答

相關問題