Question

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

• A. infinite
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is B $1$

$(k+1)x+8y=4k...\left(1\right)$
$kx+(k+3)y=3k-\mathrm{1...}\left(2\right)$
$\frac{k+1}{k}=\frac{8}{k+3}\ne \frac{4k}{3k-1}$
Now, $\frac{k+1}{k}=\frac{8}{k+3}$
$\Rightarrow k^2-4k+3=0$
$\Rightarrow k=1,3$
If $k=1$
$\frac{8}{1+3}=\frac{4}{3-1}$
$\Rightarrow 2=2$
$\therefore k=1$ is not the correct value
So, if $k=3$
$\frac{8}{3+3}\ne \frac{12}{9-1}$
$\Rightarrow \frac{4}{3}\ne \frac{3}{2}$
$\therefore k=3$ is the only value for which system of equations has no solution
Hence, number of values of $k$ is 1.