Mathematics - Determinant Question with Solution | TestHub

For $a_{1}x+b_{1}y=c_1\mathrm{and}a2x+b_{2}y=c_2$ to have no solution,

$\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$

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

$\left(k+1\right)\left(k+3\right)=8k\Rightarrow k^2-4k+3=0$

$\Rightarrow k=1,3$

but, for $k=1,\frac{8}{k+3}=\frac{4k}{3k-1}$

$\therefore k=3$

Hence, only one value