Question

A: The number of values of k for which the
equations. $(k+1)x+8y=4k$ and $kx+(k+3)y=3k-1$ has
infinitely many solution is 1
R : Two linear equations, $a_{1}x+b_{1}y+c_1=0$ ,
$a_{2}x+b_{2}y+c_2=0$ has infinite numbers solution if
det $\left[\begin{array}{cc}{a}_1& b_2\\ a_2& b_2\end{array}\right]$ = 0
The correct answer is

• A. Both A and R are true R is correct explanation to A
• B. Both A and R are true but R is not correct explanation to A
• C. A is true R is false
• D. A is false R is true

Answer 1

The correct option is C A is true R is false

$(k+1)x+8y=4k$
$kx+(k+3)y=3(k-1)$
$\Delta =\left|\begin{array}{cc}\frac{k+1}{4k}& \frac{8}{4k}\\ \frac{k}{3(k-1)}& \frac{k+3}{3(k-1)}\end{array}\right|$
$\Delta =\frac{(k+1)(k+3)}{12k(k-1)}-\frac{2k}{3k(k-1)}=0$
$\frac{(k+1)(k+3)}{12k(k-1)}=\frac{2}{3(k-1)}$
$k^2+4k+3=8k$
$k^2-4k+3=0$
$k=1,3$
but $k\ne 1$
$\therefore$ only one solution k=3