Question

Find the value of k for which the following system of equations has a unique solution:

kx + 3y = (k – 3), 12x + ky = k

Answer 1

The given system of equations:

kx + 3y = (k – 3)

⇒ kx + 3y – (k – 3) = 0   ...(i)

And 12x + ky = k

⇒ 12x + ky – k = 0   ...(ii)

These equations are of the following form:

a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0

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

For a unique solution, we must have:

`(a_1)/(a_2) ≠ (b_1)/(b_2)`

i.e., `k /12 ≠ 3/k`

⇒ k² ≠ 36 

⇒ k ≠ ±6

Thus, for all real values of k, other than ±6, the given system of equations will have a unique solution.