Question

For which value(s) of k will the pair of equations kx + 3y = k – 3, 12x + ky = k have no solution?

Answer 1

The given pair of linear equations is

kx + 3y = k – 3   ......(i)

12x + ky = k   ......(ii)

On comparing the equations (i) and (ii) with ax + by = c = 0,

We get,

a₁ = k, b₁ = 3, c₁ = –(k – 3)

a₂ = 12, b₂ = k, c₂ = – k

Then,

`a_1/a_2 = k/12`

`b_1/b_2 = 3/k`

`c_1/c_2 = (k - 3)/k`

For no solution of the pair of linear equations,

`a_1/a_2 = b_1/b_2 ≠ c_1/c_2`

`k/12 = 3/k ≠ (k - 3)/k`

Taking first two parts, we get

`k/12 = 3/k`

k² = 36

k = ± 6

Taking last two parts, we get

`3/k ≠ (k - 3)/k`

3k ≠ k(k – 3)

k² – 6k ≠ 0

So, k ≠ 0, 6

Therefore, value of k for which the given pair of linear equations has no solution is k = – 6.