Ex 3.5, 1 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 9

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Ex 3.5, 1 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 10

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Ex 3.5, 1 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 11 Ex 3.5, 1 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 12

  1. Chapter 3 Class 10 Pair of Linear Equations in Two Variables (Term 1)
  2. Serial order wise

Transcript

Ex 3.5 ,1 Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using cross multiplication method (iv) x – 3y – 7 = 0 3x – 3y – 15 = 0 x – 3y – 7 = 0 3x – 3y – 15 = 0 x – 3y – 7 = 0 Comparing with a1x + b1y + c1 = 0 ∴ a1 = 1, b1 = –3, c1 = –7 3x – 3y – 15 = 0 Comparing with a2x + b2y + c2 = 0 ∴ a2 = 3, b2 = –3, c2 = –15 a1 = 1, b1 = –3, c1 = –7 & a2 = 3, b2 = –3, c2 = –15 𝒂𝟏/𝒂𝟐 𝑎1/𝑎2 = 1/3 𝒃𝟏/𝒃𝟐 𝑏1/𝑏2 = (−3)/(−3) 𝑏1/𝑏2 = 1 𝒄𝟏/𝒄𝟐 𝑐1/𝑐2 = (−7)/(−15) 𝑐1/𝑐2 = 7/15 Since 𝒂𝟏/𝒂𝟐 ≠ 𝒃𝟏/𝒃𝟐 We have a unique solution Solving x – 3y – 7 = 0 …(1) 3x – 3y – 15 = 0 …(2) Using cross-multiplication 𝑥/(−3 ×(−15) − (−3) ×(−7) ) = 𝑦/(3 ×(−7) − 1 × (−15) ) = 1/(1 ×(−3) − 3 ×(−3) ) 𝑥/((45) − 21 ) = 𝑦/(−21 + 15 ) = 1/(−3 + 9 ) 𝑥/(24 ) = 𝑦/(−6 ) = 1/6 Now, 𝒙/𝟐𝟒 = 𝟏/𝟔 x = 24/6 ∴ x = 4 𝒚/(−𝟔) = 𝟏/𝟔 y = (−6)/6 ∴ y = – 1 Therefore, x = 4, y = –1 is the solution of our equation

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.