For which value (s) of k will the pair of equations

kx + 3y = k - 3

12x + ky = k

has no solution?

The given pair of linear equations is

kx + 3y = k - 3 …(i)

and 12x + ky = k …(ii)

On comparing with ax + by = c = 0, we get

a₁ = , b₁ = 3, c₁ = -(k - 3)

And a₂ = 12, b₂ = , c₂ = - k

a₁ /a₂ = /12

b₁ /b₂ = 3/

c₁ /c₂ =

For no solution of the pair of linear equations,

a₁/a₂ = b₁/b₂ c₁/c₂

So =

Taking first two parts, we get

/12 = 3/

k² = 36

k = 6

Taking last two parts, we get

3k k(k - 3)

k² - 6k 0

so, k 0,6

Hence, required value of k for which the given pair of linear equations has no solution is k = - 6.