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. $1$
• B. $2$
• C. $3$
• D. infinite

Answer 1

Answer: (A)

$1$

For no solution
$\frac{k+1}{k}=\frac{8}{k+3}\ne \frac{4k}{3k-1}\cdots$ (1)

$\Rightarrow (k+1)(k+3)-8k=0$
or $k^2-4k+3=0=k=1,3$
But for $k=1$, equation (1) is not satisfied Hence $k=3$.