Answer :

Given: kx + 3y = (k – 3) – eq 1

12x + ky = k – eq 2

Here,

a1 = k, b1 = 3, c1 = k – 3

a2 = 12, b2 = k, c2 = k

Given systems of equations has a unique solution

K2 ≠36

k ≠ √36

k ≠ ±6

k ≠ 6 and k ≠ – 6

That is k can be any real number other than - 6 and 6

k is any real number other than 6 and - 6

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