Question

The system $x-2y=3\mathrm{and}3x+ky=1$ has a unique solution only when

(a) k = −6
(b) k ≠ −6
(c) k = 0
(d) k ≠ 0

Answer 1

The correct option is (b).

The given system of equations can be written as follows:
x − 2y − 3 = 0 and 3x + ky − 1 = 0
The given equations are of the following form:
a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0
Here, a₁ = 1, b₁ = −2, c₁ = −3 and a₂ = 3, b₂ = k and c₂ = −1
∴ $\frac{a_1}{a_2}=\frac{1}{3},\frac{b_1}{b_2}=\frac{-2}{k}\mathrm{and}\frac{c_1}{c_2}=\frac{-3}{-1}=3$
These graph lines will intersect at a unique point when we have:
$\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$ ⇒ $\frac{1}{3}\ne \frac{-2}{k}\Rightarrow k\ne -6$
Hence, k has all real values other than −6.