Question

Find the value of k for which each of the following systems 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:
$\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$
i.e. $\frac{k}{12}\ne \frac{3}{k}$
$\Rightarrow k^2\ne 36\Rightarrow k\ne \pm 6$
Thus, for all real values of k other than $\pm 6$, the given system of equations will have a unique solution.