Ex 3.3, 1 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 14

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Ex 3.3, 1 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 15

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Ex 3.3, 1 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 16

 

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

Transcript

Ex 3.3, 1 Solve the following pair of linear equations by the substitution method. (v) √2 π‘₯+√3 𝑦=0 √3 π‘₯βˆ’βˆš8 𝑦=0 √2 π‘₯+√3 𝑦=0 √3 π‘₯βˆ’βˆš8 𝑦=0 From (1) √2 π‘₯+√3 𝑦=0 √2 π‘₯=βˆ’βˆš3 𝑦 x = (βˆ’βˆšπŸ‘ π’š)/√𝟐 Substituting x in (2) √3 π‘₯βˆ’βˆš8 𝑦=0 √3 ((βˆ’βˆš3 𝑦)/√2)βˆ’βˆš8 𝑦 = 0 (√3 Γ— (βˆ’βˆš3 𝑦))/√2βˆ’βˆš8 𝑦=0 (βˆ’3𝑦)/√2βˆ’βˆš8 𝑦=0 –3y – √8 𝑦 Γ— √2= √2 Γ— 0 –3y – √(8 Γ— 2) 𝑦=0 –3y – √16 𝑦=0 –3y – √(4 Γ— 4) 𝑦=0 –3y – √42 𝑦=0 –3y – 4y = 0 –7y = 0 y = 0/(βˆ’7) y = 0 Putting the value of y in (1) √2 π‘₯+√3 𝑦=0 √2 π‘₯+√3 Γ— 0=0 √2 π‘₯+0=0 √2 π‘₯=0 x = 0/√2 x = 0 Hence, x = 0, y = 0

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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.