The pair of linear equations 3x + 5y = 3 and 6x + ky = 8 do not have a solution if

(a) K = 5   (b) K = 10   (c) k ≠ 10   (d) k ≠ 5

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Question 24 The pair of linear equations 3x + 5y = 3 and 6x + ky = 8 do not have a solution if (a) K = 5 (b) K = 10 (c) k ≠ 10 (d) k ≠ 5 3x + 5y − 3 = 0 Comparing with a1x + b1y + c1 = 0 ∴ a1 = 3 , b1 = 5 , c1 = −3 6x + ky − 8 = 0 Comparing with a2x + b2y + c2 = 0 ∴ a2 = 6 , b2 = k, c2 = −8 Therefore, a1 = 3 , b1 = 5 , c1 = −3 & a2 = 6 , b2 = k, c2 = −8 𝑎1/𝑎2 = 3/6 𝑎1/𝑎2 = 𝟏/𝟐 𝑏1/𝑏2 = 𝟓/𝒌 𝑐1/𝑐2 = (−8)/(−3) 𝑐1/𝑐2 = 𝟖/𝟑 Since the lines have no solution 𝒂𝟏/𝒂𝟐 = 𝒃𝟏/𝒃𝟐 ≠ 𝒄𝟏/𝒄𝟐 1/2=5/𝑘 ≠ 8/3 Thus, 1/2=5/𝑘 k = 5 × 2 = 10 So, the correct answer is (b)

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo