Question

Find the value of k for which each of the following system of equations has no solution:
$3x-y-5=0,6x-2y+k=0,(k\ne 0).$

Answer 1

The given system of equations:
3x − y − 5 = 0 ...(i)
And, 6x − 2y + k = 0 ...(ii)
These equations are of the following form:
a₁x + b₁y + c₁ = 0, a₂x + b₂y + c₂ = 0
Here, a₁ = 3, b₁= −1, c₁ = −5 and a₂ = 6, b₂ = −2, c₂ = k
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. $\frac{3}{6}=\frac{-1}{-2}\ne \frac{-5}{k}$
⇒ $\frac{-1}{-2}\ne \frac{-5}{k}\Rightarrow k\ne -10$
Hence, equations (i) and (ii) will have no solution if $k\ne -10$.