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

For no solution, 

$\frac{k+1}{k}=\frac{8}{k+3}\ne \frac{4k}{3k-1}\to \left(1\right)$

$\Rightarrow (k+1)(k+3)=8k$

$\Rightarrow k=1,3$

But for $k=1,$equation (1) is not satisfied 

        Hence$k=3$