Solve linear equations by elimination and substitution x/2 + 2y/3 = -1 - Ex 3.3

part 2 - Ex 3.3, 1 (iv) - Ex 3.3 - Serial order wise - Chapter 3 Class 10 Pair of Linear Equations in Two Variables
part 3 - Ex 3.3, 1 (iv) - Ex 3.3 - Serial order wise - Chapter 3 Class 10 Pair of Linear Equations in Two Variables

  part 4 - Ex 3.3, 1 (iv) - Ex 3.3 - Serial order wise - Chapter 3 Class 10 Pair of Linear Equations in Two Variables

part 5 - Ex 3.3, 1 (iv) - Ex 3.3 - Serial order wise - Chapter 3 Class 10 Pair of Linear Equations in Two Variables part 6 - Ex 3.3, 1 (iv) - Ex 3.3 - Serial order wise - Chapter 3 Class 10 Pair of Linear Equations in Two Variables

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Ex 3.3, 1 (Elimination) Solve the following pair of linear equations by the elimination method and the substitution method : (iv) 𝑥/2+2𝑦/3=−1 𝑎𝑛𝑑 𝑥−𝑦/3=3 Given x/2+2y/3=−1 (3(x) + 2(2y))/(2 × 3)=−1 (3𝑥 + 4y)/6=−1 3x + 4y = −1 × 6 3x + 4y = −6 x – y/3=3 (3𝑥 − 𝑦 )/3=3 3x – y = 3(3) 3x – y = 9 We use elimination method with equation (1) & (2) 5y = – 15 y=(−15)/( 5) y = −3 Putting y = −3 in equation (2) 3x – y = 9 3x – (−3) = 9 3x + 3 = 9 3x = 9 – 3 3x = 6 x = 6/3 x = 2 So, x = 2, y = −3 is the solution of the given equations Ex 3.3, 1 (Substitution) Solve the following pair of linear equations by the elimination method and the substitution method : (iv) 𝑥/2+2𝑦/3=−1 𝑎𝑛𝑑 𝑥−𝑦/3=3 Given x/2+2y/3=−1 (3(x) + 2(2y))/(2 × 3)=−1 (3𝑥 + 4y)/6=−1 3x + 4y = −1 × 6 3x + 4y = −6 x – y/3=3 (3𝑥 − 𝑦 )/3=3 3x – y = 3(3) 3x – y = 9 From (1) 3x + 4y = –6 3x = –6 – 4y Putting value of 3x in (2) 3x – y = 9 (–6 – 4y) – y = 9 – 4y – y = 9 + 6 –5y = 15 y = (−15)/( 5) y = –3 Putting y = −3 in equation (2) 3x – y = 9 3x – (−3) = 9 3x = 9 – 3 3x = 6 x = 6/3 x = 2 So, x = 2, y = −3 is the solution of the given equations

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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