Here's a C program for solving simultaneous equation using Gauss elimination method using for loops with output. This program uses Array, For Loops and Nested Loops.

Enter the number of equations : 3

Enter the co-efficients of the equations :

a[1][1] = 10

a[1][2] = 1

a[1][3] = 1

b[1] = 12

a[2][1] = 2

a[2][2] = 10

a[2][3] = 1

b[2] = 13

a[3][1] = 1

a[3][2] = 1

a[3][3] = 5

b[3] = 7

The result is :

x[1] = 1.00

x[2] = 1.00

x[3] = 1.00

# include <stdio.h> # include <conio.h> void main() { int i, j, k, n ; float a[20][20], x[20] ; double s, p ; clrscr() ; printf("Enter the number of equations : ") ; scanf("%d", &n) ; printf("\nEnter the co-efficients of the equations :\n\n") ; for(i = 0 ; i < n ; i++) { for(j = 0 ; j < n ; j++) { printf("a[%d][%d] = ", i + 1, j + 1) ; scanf("%f", &a[i][j]) ; } printf("b[%d] = ", i + 1) ; scanf("%f", &a[i][n]) ; } for(k = 0 ; k < n - 1 ; k++) { for(i = k + 1 ; i < n ; i++) { p = a[i][k] / a[k][k] ; for(j = k ; j < n + 1 ; j++) a[i][j] = a[i][j] - p * a[k][j] ; } } x[n-1] = a[n-1][n] / a[n-1][n-1] ; for(i = n - 2 ; i >= 0 ; i--) { s = 0 ; for(j = i + 1 ; j < n ; j++) { s += (a[i][j] * x[j]) ; x[i] = (a[i][n] - s) / a[i][i] ; } } printf("\nThe result is :\n") ; for(i = 0 ; i < n ; i++) printf("\nx[%d] = %.2f", i + 1, x[i]) ; getch() ; }

**Output of above program is**Enter the number of equations : 3

Enter the co-efficients of the equations :

a[1][1] = 10

a[1][2] = 1

a[1][3] = 1

b[1] = 12

a[2][1] = 2

a[2][2] = 10

a[2][3] = 1

b[2] = 13

a[3][1] = 1

a[3][2] = 1

a[3][3] = 5

b[3] = 7

The result is :

x[1] = 1.00

x[2] = 1.00

x[3] = 1.00

Hello There,

Brilliant article, glad I slogged through the C Programming Tutorial it seems that a whole lot of the details really come back to from my past project.

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