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C: System Of Linear Equations In Two Variables

Tags: scanfsf
This program will demonstrate how to solve a system of linear equations in two variables. We will use Cramer's rule for this.


Let us begin with the headers.
//WinterNurf Productions

#include <stdio.h>

#include <conio.h>

Next, we declare all the required variables after beginning the main() function.
intmain()
{
      float A1, B1, C1;
      float A2, B2, C2;

      float x, y;

Next, we let the user know the format of the equations.
      printf("A1(x) + B1(y) = C1\n");

      printf("A2(x) + B2(y) = C2\n\n");

Now, we ask the user to enter all the required variables. We have used the scanf_s() function for this, but feel free to use scanf() instead.
      printf("A1 = ");
      scanf_s("%f", &A1);
      printf("B1 = ");
      scanf_s("%f", &B1);
      printf("C1 = ");
      scanf_s("%f", &C1);

      printf("\nA2 = ");
      scanf_s("%f", &A2);
      printf("B2 = ");
      scanf_s("%f", &B2);
      printf("C2 = ");

      scanf_s("%f", &C2);

If the equations do not have a unique solution, we print an error message.
      if (((A1 * B2) - (B1 * A2)) == 0) {
            printf("\nThe equatons do not have a unique soluton");

      }
If the equations do have a unique solution, we calculate it using Cramer's rule and display the result.
      else {
            x = ((C1 * B2) - (B1 * C2)) / ((A1 * B2) - (B1 * A2));
            y = ((A1 * C2) - (C1 * A2)) / ((A1 * B2) - (B1 * A2));

            printf("\nx = %f\n", x);
            printf("y = %f", y);

      }

Finally, we end the program.
      _getch();
      return 0;

}

Here is the full code.
//WinterNurf Productions

#include <stdio.h>
#include <conio.h>

intmain()
{
      float A1, B1, C1;
      float A2, B2, C2;
      float x, y;

      printf("A1(x) + B1(y) = C1\n");
      printf("A2(x) + B2(y) = C2\n\n");

      printf("A1 = ");
      scanf_s("%f", &A1);
      printf("B1 = ");
      scanf_s("%f", &B1);
      printf("C1 = ");
      scanf_s("%f", &C1);

      printf("\nA2 = ");
      scanf_s("%f", &A2);
      printf("B2 = ");
      scanf_s("%f", &B2);
      printf("C2 = ");
      scanf_s("%f", &C2);

      if (((A1 * B2) - (B1 * A2)) == 0) {
            printf("\nThe equatons do not have a unique soluton");
      }
      else {
            x = ((C1 * B2) - (B1 * C2)) / ((A1 * B2) - (B1 * A2));
            y = ((A1 * C2) - (C1 * A2)) / ((A1 * B2) - (B1 * A2));

            printf("\nx = %f\n", x);
            printf("y = %f", y);
      }

      _getch();
      return 0;
}

This program has been written in Microsoft Visual Studio Professional 2013. It may not work in older compilers. However, simple editing will make the code compatible.


This post first appeared on Code Walk, please read the originial post: here

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C: System Of Linear Equations In Two Variables

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