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

• A. \N
• B. 1
• C. 2
• D. Infinite

Answer 1

The correct option is B 1

For infinitely many solutions, the two equations must be identical
$\Rightarrow \frac{k+1}{k}=\frac{8}{k+3}=\frac{4k}{3k-1}\Rightarrow (k+1)(k+3)=8k$ and 8(3k-1) = 4k(k+3)
$\Rightarrow k^2-4k+3=0$ and $k^2-3k+2=0.$
By cross multiplication, $\frac{k^2}{-8+9}=\frac{k}{3-2}=\frac{1}{-3+4}$
$k^2=1$ and k=1 ;$\therefore$ k=1.