Question

For what value of k, the system of equations
$x+2y=5,3x+ky+15=0$
has (i) a unique solution, (ii) no solution?

Answer 1

The given system of equations is:
x + 2y = 5
⇒ x + 2y − 5= 0 ...(i)
3x + ky + 15 = 0 ...(ii)
These equations are of the form:
a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0
where, a₁ = 1, b₁= 2, c₁ = −5 and a₂ = 3, b₂ = k, c₂ = 15
(i) For a unique solution, we must have:
$\frac{a_1}{a_2}\ne \frac{b_1}{b_2}$, i.e., $\frac{1}{3}\ne \frac{2}{k}\Rightarrow k\ne 6$
Thus, for all real values of k other than 6, the given system of equations will have a unique solution.
(ii) For the given system of equations to have no solutions, we must have:
$\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$
$\Rightarrow \frac{1}{3}=\frac{2}{k}\ne \frac{-5}{15}$
$\Rightarrow \frac{1}{3}=\frac{2}{k}\mathrm{and}\frac{2}{k}\ne \frac{-5}{15}$
$\Rightarrow k=6,k\ne -6$
Hence, the required value of k is 6.