Solve the system of two equations with two unknowns below:
a1, b1, c1, a2, b2 and c2 are inputted by the user himself.
I've been trying to find a math solution for the problem first and I can't seem to go far..
What I've tried so far is :
Is this correct or is there an easier way to do this?
Current code:
#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;
}
In order to solve for a given number of unknowns, we require that the same number of equations be provided. For instance, we would require two equations to solve for two unknown quantities. We would require three equations to solve for three unknown quantities, and so on.
we solve the linear system using Cramer's rule:
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 equations system: result, x = 1.000000, y = 2.000000
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