#### 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)

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 the given linear equations,

a

_{1}/a

_{2}= b

_{1}/b

_{2}≠ c

_{1}/c

_{2}

k/12 = 3/k ≠ (k-3)/k

Taking the given first two parts, we have,

k/12 = 3/k

k

^{2}= 36

k = + 6 ,-6

Taking the given last two parts, we have,

3/k ≠ (k-3)/k

3k ≠ k(k – 3) k

^{2}– 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.