Question

Find the value of k for which 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

where, a₁ = k, b₁ = 3, c₁ = –3 and a₂ = 12, b₂ = k, c₂ = –6

In order that the given system has no solution, we must have:

`(a_1)/(a_2) = (b_1)/(b_2) ≠ (c_1)/(c_2)`

i .e., `k/12 = 3/k ≠ (-3)/(-6)`

`k/12 = 3/k` and `3/k ≠ 1/2`

⇒ k² = 36 and k ≠ 6

⇒ k = ±6 and k ≠ 6

Hence, the given system of equations has no solution when k is equal to –6.