The system x – 2y = 3 and 3x + ky = 1 has a unique solution only when

We have,

x – 2y – 3 = 0

3x + ky – 1 = 0

The given equation is in the form: a₁x ₊ b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0

Here, we have:

a₁ = 1, b₁ = – 2 and c₁ = – 3

And, a₂ = 3, b₂ = k and c₂ = – 1

∴ , and

These graph lines will intersect at a unique point when we have:

∴

Hence, k has all real values other than – 6

Thus, option B is correct