The number of values of $k$ , for which the system of equations  $\left(k+1\right)x+8y=4k;$  $kx+\left(k+3\right)y=3k-1$  has infinitely ,many solutions is

The given system of equations are $\left(k+1\right)x+8y=4k$  and

$kx+\left(k+3\right)y=3k-1$

Given that, the system of equations has infinitely many solutions,

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

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

After simplification we get $k=1$