Question

Find the value of k for which each of the following systems of equations has a unique solution:
$kx+2y=5,3x+y=1.$

Answer 1

The given system of equations:
kx + 2y = 5
⇒ kx + 2y − 5 = 0 ....(i)
And, 3x + y = 1
⇒ 3x + y − 1 = 0 ....(ii)
These equations are of the following form:
a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0
Here, a₁ = k, b₁= 2, c₁ = −5 and a₂ = 3, b₂ = 1, c₂ = −1
For a unique solution, we must have:
$\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$ i.e. $\frac{k}{3}\ne \frac{2}{1}$
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.