The number of values of k for which the system of equations: $(k+1)x+8y = 4k, kx+(k+ 3) y = 3k -1$ has no solution, is
Answer & explanation
Correct answer: option 2
The given system of equations will have no solution if
$\frac{k+1}{k}=\frac{8}{k+3}≠\frac{4k}{3k-1}$
$⇒\frac{k+1}{k}=\frac{8}{k+3}$ and $\frac{8}{k+3}≠\frac{4k}{3k-1}$
$⇒k^2 + 4k+3=8k$ and $24k -8 ≠4k^2 + 12k$
$⇒k^2 - 4k+3=0$ and $4k^2 -12k + 8≠0$
$⇒k =1, 3$ and $k^2 - 3k+2≠0⇒ k = 3$
Hence, there is only one value of k.