Question

Find the value of k for which the system of equations $3x+y=1\mathrm{and}kx+2y=5$ has (i) a unique solution, (ii) no solution.

Answer 1

The given system of equations:
3x + y = 1
⇒ 3x + y − 1= 0 ....(i)
kx + 2y = 5
⇒ kx + 2y − 5 = 0 ....(ii)
These equations are of the following form:
a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0
Here, a₁ = 3, b₁= 1, c₁ = −1 and a₂ = k, b₂ = 2, c₂ = −5

(i) For a unique solution, we must have:
$\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$ i.e. $\frac{3}{k}\ne \frac{1}{2}\Rightarrow k\ne 6$
Thus, for all real values of k other than 6, the given system of equations will have a unique solution.

(ii) In order that the given equations have no solution, we must have:
$\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$
$\frac{3}{k}=\frac{1}{2}\ne \frac{-1}{-5}$
$\Rightarrow \frac{3}{k}=\frac{1}{2}\mathrm{and}\frac{3}{k}\ne \frac{-1}{-5}$
$\Rightarrow k=6,k\ne 15$
Thus, for k = 6, the given system of equations will have no solution.