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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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.