Question :

For which value(s) of k would the pair of given equations kx + 3y = k – 3, 12x + ky = k have no solution?
Submitted on 15/5/2024 | Answered by Vandana Rana

Answer :

The given pair of the linear equations is as follows:

kx + 3y = k – 3 …(i)
12x + ky = k …(ii)

a1 /a2 = k/12
b1/b2= 3/k
c1/c2 = (k-3)/k

For no solution of the pair of the given linear equations,

a1/a2 = b1/b2 ≠ c1/c2

k/12 = 3/k ≠ (k-3)/k

Taking the given first two parts, we have,

k/12 = 3/k
k2= 36
k = + 6 ,-6

Taking the given last two parts, we have,

3/k ≠ (k-3)/k
3k ≠ k(k – 3)
k2 – 6k ≠ 0
so, k ≠ 0,6

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