kx + 3y – (k – 3) = 0, 12x + ky – k = 0

Question

Find the values of a and b for which the following system of linear equations has infinitely many solutions:

kx + 3y – (k – 3) = 0, 12x + ky – k = 0

Answer 1

Given, pair of equations

kx + 3y – (k – 3) = 0

and 12x + ky – k = 0

On comparing the given equation with standard form i.e. a₁ x + b₁y + c₁ = 0 and a₂ x + b₂y + c₂ = 0, we get

a₁ = k, b₁ = 3 and c₁ = – (k – 3)

and a_{2 =} 12, b₂ = k and c₂ = – k

For infinitely many solutions,

⇒ k² = 36

⇒ k = √36

⇒ k = ±6

But k = – 6 not satisfies the last two terms of eqⁿ (1)

On taking II and III terms, we get

3/k = (k -3)/k

⇒ 3k = k(k – 3)

⇒ 3k = k^{2 –} 3k

⇒ k² – 3k – 3k = 0

⇒ k(k – 6) = 0

⇒ k = 0 and 6

Which satisfies the last two terms of eqⁿ (1)

Hence, the required value of k = 0, 6