Question

Find the value of k for which each of the following systems of equations has a unique solution:
$x-2y=3,3x+ky=1.$

Answer 1

The given system of equations:
x − 2y = 3
⇒ x − 2y − 3 = 0 ....(i)
And, 3x + ky = 1
⇒ 3x + ky − 1 = 0 ....(ii)
These equations are of the following form:
a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0
Here, a₁ = 1, b₁= −2, c₁ = −3 and a₂ = 3, b₂ = k, c₂ = −1
For a unique solution, we must have:
$\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$ i.e. $\frac{1}{3}\ne \frac{-2}{k}$
This happens when k ≠ −6.
Thus, for all real values of k other than −6, the given system of equations will have a unique solution.