The number of values of k for which the system of equations

$(k + 1) x + 8y = 4k$

$kx+(k+ 3) y = 3k -1$

has infinitely many solutions, is

1

The given system of equations will have infinitely many solutions, if

$\begin{vmatrix}k+1&8\\k&k+3\end{vmatrix}=0,\begin{vmatrix}k+1&4k\\k&3k-1\end{vmatrix}=0$ and $\begin{vmatrix}4k&8\\3k-1&k+3\end{vmatrix}=0$

$⇒ k^2-4k+3=0,-k^2 + 2k-1=0$ and $4k^2 - 12k +8=0$

$⇒ (k-1) (k − 3) = 0, (k-1)^2 = 0, (k-1) (k − 2) = 0$

$⇒ k=1$