The number of values of ‘k’ for which the system of equations k+1x+8y=4k, kx+(k+3)y=3k−1 has no solution, is

As the system of equations has no solution⇒k+1k=8k+3≠4k3k-1 Now, (k+1)(k+3)=8k⇒k2+4k+3=8k⇒k2-4k+3=0⇒k=1,3But for k=1, the inequality is not satisfied. ⇒k=3 is the only value.