Question

The value of k for which kx+3y-k+3=0 and 12x+ky=k have infinitely many solutions is :

• A. 1
• B. 5
• C. -6
• D. 6

Answer 1

Answer:

D. 6

$kx+3y-(k-3)=0$

$12x+ky-k=0$

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

$\frac{k}{12}=\frac{3}{k}=\frac{k-3}{k}$

$\frac{k}{12}=\frac{3}{k}$

$k^2=36$

$k=±6$

$k^2-3k=3k$

$k^2-6k=0$

$k(k-6)=0$

$k=0$

$k-6=0$

$k=6$