Question

If the system of equations $kx+3y-(k-3)=0,12x+ky-k=$ has infinitely many solutions, then $k=$

• A. $6$
• B. $-6$
• C. $0$
• D. None of these

Answer 1

The correct option is A $6$

For infinitely many solution
$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
$\frac{K}{12}=\frac{3}{K}=\frac{-(K-3)}{-K}$
By condition (i) and (ii)
$K^2=36\Rightarrow K=\pm 6$ ...(i)
By, condition (i) and (ii)
$3K=K^2-3K$
$K^2-6K=0$
$K(K-6)=0$
$K=0$ or $K=6$ ...(ii)
So, by equation (i) and (ii)
$K=6$