#include <stdio.h>

void solution( int a[][20], int var );
int main()
{

int a[ 20 ][ 20 ], var, i, j, k, l, n;
printf( "\nEnter the number of variables:\n" );
scanf( "%d", &var );

for ( i = 0;i < var;i++ )
{
printf( "\nEnter the equation%d:\n", i + 1 );

for ( j = 0;j < var;j++ )
{
printf( "Enter the coefficient of x%d:\n", j + 1 );
scanf( "%d", &a[ i ][ j ] );
}

printf( "\nEnter the constant:\n" );
scanf( "%d", &a[ i ][ var] );
}

solution( a, var );
return 0;
}



void solution( int a[ 20 ][ 20 ], int var )
{
int k, i, l, j;

for ( k = 0;k < var;k++ )
{
for ( i = 0;i <= var;i++ )
{
l = a[ i ][ k ];

for ( j = 0;j <= var;j++ )
{
if ( i != k )
a[i][j] = (a[k][k]*a[i][j])-(l*a[k][j]);
}
}
}

printf( "\nSolutions:" );

for ( i = 0;i < var;i++ )
{
printf( "\nTHE VALUE OF x%d IS %f\n", i + 1, ( float ) a[ i ][ var ] / ( float ) a[ i ][ i ] );
}

}

Output

$ gcc bubblesort.c -o bubblesort
$ ./bubblesort

Enter the number of variables: 3

Enter the equation 1:
Enter the coefficient of x1: 1
Enter the coefficient of x2: 0
Enter the coefficient of x3: 0

Enter the constant: 2

Enter the equation 2:
Enter the coefficient of x1: 0
Enter the coefficient of x2: 1
Enter the coefficient of x3: 0

Enter the constant: 0

Enter the equation 3:
Enter the coefficient of x1: 0
Enter the coefficient of x2: 0
Enter the coefficient of x3: 1

Enter the constant: -1

Solutions:
THE VALUE OF x1 IS 2.000000

THE VALUE OF x2 IS 0.000000

THE VALUE OF x3 IS -1.000000