Last updated at Dec. 13, 2024 by Teachoo
Question 6 For what values of k will the following pair of linear equations have infinitely many solutions? kx + 3y – (k-3) = 0 12x + ky – k = 0 kx + 3y – (k – 3) = 0 12x + ky – k = 0 kx + 3y – (k – 3) = 0 Comparing with a1x + b1y + c1 = 0 ∴ a1 = k , b1 = 3 , c1 = – (k – 3) 12x + ky – k = 0 Comparing with a2x + b2y + c2 = 0 ∴ a2 = 12 , b2 = k , c2 = –k Since equation has infinite number of solutions So, 𝑎1/𝑎2 = 𝑏1/𝑏2 = 𝑐1/𝑐2 Putting values 𝑘/12 = 3/𝑘 = (−(𝑘 − 3))/(−𝑘) 𝑘/12 = 3/𝑘 = (𝑘 − 3)/𝑘 Solving 𝒌/𝟏𝟐 = 𝟑/𝒌 k2 =12 × 3 k2 = 36 k = ±√36 k = ± √62 k = ± 6 So, k = 6, – 6 Solving 𝟑/𝒌 = (𝒌 −𝟑)/𝒌 3k = k(k -3) 3k = k2 – 3k 0 = k2 – 3k – 3k k2 – 3k – 3k = 0 k2 – 6k = 0 k(k – 6) = 0 So, k = 0 , k = 6 Since k = 6 satisfies both equations. Hence, k = 6 is the answer
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About the Author
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