Ex 3.3, 1 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 17

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Ex 3.3, 1 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 18

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Ex 3.3, 1 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 19 Ex 3.3, 1 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 20

 

  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. (vi) 3π‘₯/2βˆ’5𝑦/3=βˆ’2 π‘₯/3+𝑦/2=13/6 Removing fractions from both equations πŸ‘π’™/πŸβˆ’πŸ“π’š/πŸ‘=βˆ’2 Multiplying both equations by 6 6 Γ— 3π‘₯/2βˆ’"6 Γ—" 5𝑦/3="6 Γ—"βˆ’2 9x – 10y = βˆ’ 12 𝒙/πŸ‘+π’š/𝟐=πŸπŸ‘/πŸ” Multiplying both equations by 6 6 Γ— π‘₯/3+"6 Γ—" 𝑦/2="6 Γ— " 13/6 2x + 3y = 13 Hence, our equations are 9x – 10y = βˆ’12 …(1) 2x + 3y = 13 …(2) From (1) 9x – 10y = –12 9x = –12 + 10y x = ((βˆ’πŸπŸ + πŸπŸŽπ’š)/πŸ—) Substituting value of x in (2) 2x + 3y = 13 2((βˆ’12 + 10𝑦)/9)+3𝑦=13 Multiplying both sides by 9 9 Γ— 2((βˆ’12 + 10𝑦)/9) + 9 Γ— 3y = 9 Γ— 13 (2(βˆ’12 + 10𝑦) + 9 Γ— 3𝑦)/9=13 2(–12 + 10y) + 27y = 13Γ—9 – 24 + 20y + 27y = 117 – 24 + 20y + 27y – 117 = 0 47y – 141 = 0 47y = 141 y = 141/47 y = 3 Putting y = 3 in (2) 2x + 3y = 13 2x + 3(3) = 13 2x + 9 = 13 2x = 13 – 9 2x = 4 x = 4/2 x = 2 Hence, x = 2, y = 3 is the solution of 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.