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 infinitely many solutions is

Answer 1

Answer: 1

$(k+1)x+8y=4k$
$kx+(k+3)y=3k-1$
for these equation to have infinite solution
$D=D_1=D_2=0$
$D=\left|\begin{array}{cc}k+1& 8\\ k& k+3\end{array}\right|=k^2+4k+3-8k$
$=k^2-4k+3=0$
$\therefore k=3$ or $k=1$ ...(1)
$D_1=\left|\begin{array}{cc}4k& 8\\ 3k-1& k+3\end{array}\right|=4k^2+12k-24k+8$
$=4k^2-12k+8=0$
$\therefore k^2-3k+2=0$
$k=2$ or $k=1$ ...(2)
$D_2=\left|\begin{array}{cc}k+1& 4k\\ k& 3k-1\end{array}\right|=3k^2-k+3k-1-4k^2=0$
$k^2-2k+1=0$
$\Rightarrow k=1$ ...(3)
From (1), (2), (3) $k=1$
$\therefore$ Only one value of $k$
Alternatively: Lines must be identical
i.e. $\frac{k+1}{k}=\frac{8}{k+3}=\frac{4k}{3k-1}$
$\Rightarrow k=1$