For what value of k will the following pair of linear equations have infinite solutions?
$k x+3 y-(k-3)=0$
$12 x+k y-k=0$
Answer & explanation
Correct answer: option 2
$k x+3 y-(k-3)=0$
$12 x+k y-k=0$
a1 = k, b1 = 3, c1 = -(k-3)
a2 = 12, b2 = k and c2 = -k
For infinite solutions, a1/a2 = b1/b2 = c1/c2
k/12 = 3/k = -(k-3)/-k
$k^2 = 36$ ⇒ k = ± 6
and k-3 = 3 ⇒ k = 6
The correct answer is Option (2) → 6 Only