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 C $2$

The equations are in the form,

$a_{1}x+b_{1}y+c_1=0$

and $a_{2}x+b_{2}y+c_2=0$

As, the equations are inconsistent,.

We can say that,.

$\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$

Hence,

$\frac{k+1}{k}=\frac{8}{k+3}$

$⟹(k+1)(k+3)=8k$

$⟹k²+4k+3=8k$

$⟹k²+4k+3-8k=0$

$⟹k²-4k+3=0$

$⟹(k-3)(k-1)=0$

For the equation to be $0$,

Either,

$k-3=0$ (or) $k-1=0$

$k=3$ (or) $k=1$.

Hence the number of values of $k$ is $2$.