Question

Find the values of k for which the pair of linear equations 2x + 3y = k – 2 and 12x + ky = k has no

solution​

Answer 1

Answer:

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, a1 /a2 = k/12 b1 /b2 = 3/k c1 /c2 = (k-3)/k For no solution of the pair of linear equations, a1/a2 = b1/b2≠ c1/c2 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.Read more on Sarthaks.com - https://www.sarthaks.com/878788/for-which-value-s-ofkwill-the-pair-of-equations-kx-3y-k-3-12x-ky-k-have-no-solution

Answer 2

Answer:

2× +3y = 12 × k -2 + ky = k +

Step-by-step explanation:

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