Question

The value of k for which kx+3y-k+3=0and 12x+ky=k, have infinite solution

Answer 1

kx+3y-k+3=0 12x+ky=k

⇒ kx+3y-(k-3) =0 ⇒ 12x+ky-k =0

Here, a₁=k , b₁=3 , c₁= -(k-3) a₂=12 , b₂=k , c₂=-k

For infinite number of solutions,

a₁/a₂ = b₁/b₂ = c₁/c₂

k/12 = 3/k = -(k-3) /-k

⇒ k² =36

⇒ k = √36 = 6 , -6

∴ k = 6, -6