b
(b) For infinitely many solutions, the two equations must be identical
$ \Rightarrow \frac{{k + 1}}{k} = \frac{8}{{k + 3}} = \frac{{4k}}{{3k - 1}}$
$ \Rightarrow (k + 1)(k + 3) = 8k$ and $8\,(3k - 1) = 4k(k + 3)$
$ \Rightarrow {k^2} - 4k + 3 = 0$ and ${k^2} - 3k + 2 = 0$.
By cross multiplication, $\frac{{{k^2}}}{{ - 8 + 9}} = \frac{k}{{3 - 2}} = \frac{1}{{ - 3 + 4}}$
${k^2} = 1$ and $k = 1$; $k = 1$.