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Ex 3.5, 2 (i) For which values of a and b does the pair - Ex 3.5

Ex 3.5, 2 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 2
Ex 3.5, 2 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 3
Ex 3.5, 2 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 4

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Ex 3.5, 2 (i) For which values of a and b does the following pair of linear equations have an infinite number of solutions? 2x+ 3y = 7 (a – b) x + (a + b) y= 3a + b – 2 2x + 3y = 7 (a – b)x + (a + b) y = 3a + b – 2 2x + 3y = 7 2x + 3y – 7 = 0 Comparing with a1x + b1y + c1 = 0 ∴ a1 = 2,b1 = 3,c1 = –7 (a – b)x + (a + b)y = 3a + b – 2 (a – b)x + (a + b)y – (3a + b – 2)= 0 Comparing with a2x + b2y + c2 = 0 ∴ a2 = (a – b),b2 = (a + b), c2 = – (3a + b – 2) So, a1 = 2, b1 = 3, c1 = –7 & a2 = (a – b), b2 = (a + b), c2 = – (3a + b – 2) It is given that the equation has infinite number of solutions So, 𝒂𝟏/𝒂𝟐 = 𝒃𝟏/𝒃𝟐 = 𝒄𝟏/𝒄𝟐 Putting in values 2/((𝑎 − 𝑏)) = 3/((𝑎 + 𝑏)) = (−7)/(−(3𝑎 + 𝑏 − 2)) 𝟐/((𝒂 − 𝒃)) = 𝟑/((𝒂 + 𝒃)) = (−𝟕)/(−(𝟑𝒂 + 𝒃 − 𝟐)) Solving 𝟐/((𝒂 − 𝒃)) = 𝟑/((𝒂 + 𝒃)) 2(a + b) = 3(a – b) 2a + 2b = 3a – 3b 2b + 3b = 3a – 2a 5b = a a = 5b Solving 𝟐/((𝒂 − 𝒃)) = 𝟕/((𝟑𝒂 + 𝒃 − 𝟐)) 2(3a + b – 2) = 7(a - b) 6a + 2b – 4 = 7a – 7b 2b – 4 + 7b = 7a – 6a 9b – 4 = a Comparing (3) and (4) 5b = 9b – 4 4 = 9b – 5b 4 = 4b 4b = 4 b = 𝟒/𝟒 = 1 Putting b = 1 in (3) a = 5b a = 5(1) a = 5 Therefore, for a = 5, b = 1 the given set of equations have infinitely many solutions

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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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.