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  

Solution:

The given system of equations will have no

  solution if

 $\frac{k+1}{k}=\frac{8}{k+3}\ne \frac{4k}{3k-1}$

$\Rightarrow \frac{k+1}{k}=\frac{8}{k+3}$ and $\frac{8}{k+3}\ne \frac{4k}{3k-1}$

 $\Rightarrow k^2+4k+3=8k$ and $24k-8\ne 4k^2+12k$

$\Rightarrow k^2-4k+3=0$ and $4k^2-12k+8\ne 0$

$\Rightarrow k=1,3$ and $k^2-3k+2\ne 0\Rightarrow k=3$ 

Hence, there is only one value of  $k$