Question

Find the value of k for which the following system of linear equations has an infinite number of solutions:

(k – 3)x + 3y – k, kx + ky – 12 = 0

Answer 1

The given system of equations can be written as

(k – 3)x + 3y – k = 0

kx + ky – 12 = 0

This system is of the form:

a₁x + b₁y + c₁ = 0

a₂x + b₂y + c₂ = 0

where, a₁ = k, b₁ = 3, c₁ = –k and a₂ = k, b₂ = k, c₂ = –12

For the given system of equations to have a unique solution, we must have:

`(a_1)/(a_2) = (b_1)/(b_2) = (c_1)/(c_2)`

⇒ `(k - 3)/k = 3/k = (-k)/(-12)`

⇒ k – 3 = 3 and k² = 36

⇒ k = 6 and k = ± 6

⇒ k = 6

Hence, k = 6.