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Ex 3.4, 1 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 18

Ex 3.4, 1 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 19
Ex 3.4, 1 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 20

Ex 3.4, 1 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 21 Ex 3.4, 1 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 22 Ex 3.4, 1 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 23

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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) π‘₯/2+2𝑦/3=βˆ’1 π‘Žπ‘›π‘‘ π‘₯βˆ’π‘¦/3=3 Given x/2+2y/3=βˆ’1 (3(x) + 2(2y))/(2 Γ— 3)=βˆ’1 (3π‘₯ + 4y)/6=βˆ’1 3x + 4y = βˆ’1 Γ— 6 3x + 4y = βˆ’6 x – y/3=3 (3π‘₯ βˆ’ 𝑦 )/3=3 3x – y = 3(3) 3x – y = 9 We use elimination method with equation (1) & (2) 5y = – 15 y=(βˆ’15)/( 5) y = βˆ’3 Putting y = βˆ’3 in equation (2) 3x – y = 9 3x – (βˆ’3) = 9 3x + 3 = 9 3x = 9 – 3 3x = 6 x = 6/3 x = 2 So, x = 2, y = βˆ’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) π‘₯/2+2𝑦/3=βˆ’1 π‘Žπ‘›π‘‘ π‘₯βˆ’π‘¦/3=3 Given x/2+2y/3=βˆ’1 (3(x) + 2(2y))/(2 Γ— 3)=βˆ’1 (3π‘₯ + 4y)/6=βˆ’1 3x + 4y = βˆ’1 Γ— 6 3x + 4y = βˆ’6 x – y/3=3 (3π‘₯ βˆ’ 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 = (βˆ’15)/( 5) y = –3 Putting y = βˆ’3 in equation (2) 3x – y = 9 3x – (βˆ’3) = 9 3x = 9 – 3 3x = 6 x = 6/3 x = 2 So, x = 2, y = βˆ’3 is the solution of the given equations

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