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Ex 3.6, 1 (i) and (ii) - Solve 1/2x + 1/3y = 2 , 1/3x +1/2y

Ex 3.6, 1 (i) and (ii) - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 2
Ex 3.6, 1 (i) and (ii) - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 3
Ex 3.6, 1 (i) and (ii) - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 4

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Ex 3.6, 1 (i) and (ii) - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 5

Ex 3.6, 1 (i) and (ii) - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 6
Ex 3.6, 1 (i) and (ii) - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 7
Ex 3.6, 1 (i) and (ii) - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 8

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Transcript

Ex 3.6, 1 Solve the following pairs of equations by reducing them to a pair of linear equations: (i) 1/2π‘₯ + 1/3𝑦 = 2 1/3π‘₯ + 1/2𝑦 = 13/6 1/2π‘₯ + 1/3𝑦 = 2 1/3π‘₯ + 1/2𝑦 = 13/6 Let 1/π‘₯ = u 1/𝑦 = v So, our equations become 1/2 u + 1/3 v = 2 (3𝑒 + 2𝑣)/(2 Γ— 3) = 2 3u + 2v = 12 1/3 u + 1/2 v = 13/6 (2𝑒 +3𝑣)/(2 Γ— 3) = 13/6 2u + 3v = 13 Our equations are 3u + 2v = 12 …(3) 2u + 3v = 13 …(4) From (3) 3u + 2v = 12 3u = 12 – 2v u = (12 βˆ’ 2𝑣)/3 Putting value of u in (4) 2u + 3v = 13 2 ((12 βˆ’2𝑣)/3) + 3v = 13 Multiplying both sides by 3 3 Γ— 2((12 βˆ’ 2𝑣)/3) + 3 Γ— 3v = 3 Γ— 13 2(12 – 2v) + 9v = 39 24 – 4v + 9v = 39 – 4v + 9v = 39 – 24 5v = 15 v = 15/5 v = 3 Putting v = 3 in (3) 3u + 2v = 12 3u + 2(3) = 12 3u + 6 = 12 3u = 12 – 6 3u = 6 u = 6/3 u = 2 Hence, v = 3, u = 2 But we have to find x & y We know that u = 𝟏/𝒙 2 = 1/π‘₯ x = 𝟏/𝟐 v = 𝟏/π’š 3 = 1/𝑦 y = 𝟏/πŸ‘ So, x = 𝟏/𝟐 , y = 𝟏/πŸ‘ is the solution of the given equation Ex 3.6, 1 Solve the following pairs of equations by reducing them to a pair of linear equations: (ii) 2/√π‘₯ + 3/βˆšπ‘¦ = 2 4/√π‘₯ βˆ’ 9/βˆšπ‘¦ = –1 2/√π‘₯ + 3/βˆšπ‘¦ = 2 4/√π‘₯ βˆ’ 9/βˆšπ‘¦ = βˆ’1 So, our equations become 2u + 3v = 2 4u – 9v = –1 Our equations 2u + 3v = 2 …(3) 4u – 9v = –1 …(4) From (3) 2u + 3v = 2 2u = 2 – 3v u = (2 βˆ’ 3𝑣)/2 Putting value of u in (4) 4u – 9v = – 1 4 ((2 βˆ’ 3𝑣)/2) – 9v = –1 2(2 – 3v) – 9v = –1 4 – 6v – 9v = –1 – 6v – 9v = –1 – 4 –15v = – 5 v = (βˆ’5)/(βˆ’15) v = 𝟏/πŸ‘ Putting v = 1/3 in (3) 2u + 3v = 2 2u + 3 (1/3) = 2 2u + 1 = 2 2u = 2 – 1 u = 𝟏/𝟐 Hence, u = 1/2 & v = 1/3 But, we need to find x & y u = 𝟏/βˆšπ’™ 1/2 = 1/√π‘₯ √π‘₯ = 2 Squaring both sides (√π‘₯)2 = (2)2 x = 4 v = 𝟏/βˆšπ’š 1/3 = 1/βˆšπ‘¦ βˆšπ‘¦ = 3 Squaring both sides (βˆšπ‘¦)2 = (3)2 y = 9 Therefore, x = 4, y = 9 is the solution of the given equation

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