Question

Find the value of k for which the system of equations

kx + 3y + 3 – k = 0, 12x + ky – k = 0 has no solution.

Find the value of k for which the following system of equations has no solution:

kx + 3y = k – 3, 12x + ky = k

Answer 1

The given system of equations can be written as

kx + 3y + 3 – k = 0   ...(i)

12x + ky – k = 0   ...(ii)

This system of the form:

a₁x + b₁y + c₁ = 0

a₂x + b₂y + c₂ = 0

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

For the given system of linear equations to have no solution, we must have:

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

⇒ `k/12 = 3/k ≠ (3 - k)/(-k)`

⇒ `k/12 = 3/k` and `3/k ≠ (3 - k)/(-k)`

⇒ k² = 36 and –3 ≠ 3 – k

⇒ k = ±6 and k ≠ 6

⇒ k = –6

Hence, k = –6.