[" i."(3k+1)x+3y=2],[(k2+1)x+(k-2)y=5]

Find the value of k that the following system of linear equation has no solution (3k+1)x+3y-2=0, (k^2+1)x+(k-2)y-5=0 .

If the pair of linear equations (3k+1)x+3y-2 =0 and (k^2+1)x+(k-2)y-5=0 inconsistent, The value of k is

Determine the value of k so that the following linear equations have no solution: (3k+1)x+3y-2=0,quad (k^(2)+1)x+(k-2)y-5=0

If the pair of linear equations (3k+1)x+3y-2=0 and (k^(2)+1)x+(k-2)y-5=0 inconsistent The value of k is

Determine the value of k so that the following linear equations have no solution: (3k+1)x+3y-2=0,\ \ \ \ (k^2+1)x+(k-2)y-5=0

For what value of k, will the following pair of equations have infinitely many solutions: 2x+ 3y=7 and (k+2)x-3 (1-k) y= 5k +1

If the system of equations (k+1)^(3)x+(k+2)^(3)y=(k+3)^(2),(k+1)x+(k+2)y+k+3,x+y=1 is consistent then the value of k is

{:(2x + 3y = 2),((k + 2)x + (2k + 1)y = 2(k - 1)):}

If the system of equation (k+1)^(3)x+(k+2)^(3)y=(k+3)^(3),(k+1)x+(k+2)y=k+3,x+y=1 is consistent,then the value of k is