Question

For what value of k, the system of equations

has (i) a unique solution, (ii) no solution?

Answer 1

The given system of equations is:
kx + 2y = 5
⇒ kx + 2y − 5= 0 ...(i)
3x − 4y = 10
⇒ 3x − 4y − 10 = 0 ...(ii)
These equations are of the forms:
a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0
where, a₁ = k, b₁= 2, c₁ = −5 and a₂ = 3, b₂ = −4, c₂ = −10
(i) For a unique solution, we must have:
$\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$, i.e., $\frac{k}{3}\ne \frac{2}{-4}\Rightarrow k\ne \frac{-3}{2}$
Thus for all real values of k other than $\frac{-3}{2}$, the given system of equations will have a unique solution.
(ii) For the given system of equations to have no solutions, we must have:
$\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$
$\Rightarrow \frac{k}{3}=\frac{2}{-4}\ne \frac{-5}{-10}$
$\Rightarrow \frac{k}{3}=\frac{2}{-4}\mathrm{and}\frac{\mathrm{k}}{3}\ne \frac{1}{2}$
$\Rightarrow k=\frac{-3}{2},k\ne \frac{3}{2}$
Hence, the required value of k is $\frac{-3}{2}$.