Question

Find the value of k for which each of the following system of equations has no solution:
$kx+3y=3,12x+ky=6.$

Answer 1

The given system of equations:
kx + 3y = 3
kx + 3y − 3 = 0 ....(i)
12x + ky = 6
12x + ky − 6= 0 ....(ii)
These equations are of the following form:
a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0
Here, a₁ = k, b₁= 3, c₁ = −3 and a₂ = 12, b₂ = k, c₂ = −6
In order that the given system of equations has no solution, we must have:
$\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$
i.e. $\frac{k}{12}=\frac{3}{k}\ne \frac{-3}{-6}$
$\frac{k}{12}=\frac{3}{k}$ and $\frac{3}{k}\ne \frac{1}{2}$
$\Rightarrow k^2=36\mathrm{and}\mathrm{k}\ne 6$
$\Rightarrow k=\pm 6\mathrm{and}k\ne 6$
Hence, the given system of equations has no solution when k is equal to −6.