The number of values of k , for which the system of equations : (k...

For ${a}_{1}x + {b}_{1}y = {c}_{1}anda2x + {b}_{2}y = {c}_{2}$ to have no solution,

$\dfrac{{a}_{1}}{{a}_{2}} = \dfrac{{b}_{1}}{{b}_{2}} \ne \dfrac{{c}_{1}}{{c}_{2}}$

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

\left(k + 1\right)\left(k + 3\right) = 8k ⇒ {k}^{2} - 4k + 3 = 0$</p><p>$⇒ k = 1, 3$</p><p>but, for$k = 1, \dfrac{8}{k + 3} = \dfrac{4k}{3k - 1}$</p><p>$∴ k = 3$

Hence, only one value