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Ex 3.6, 1 (iii) & (iv) : 4/x + 3y  = 14 , 3/x - 4y = 23 - Ex 3.6

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

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

Ex 3.6, 1 (iii) and (iv) - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 6
Ex 3.6, 1 (iii) and (iv) - Chapter 3 Class 10 Pair of Linear Equations in Two Variables - Part 7
Ex 3.6, 1 (iii) and (iv) - 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: (iii) 4/π‘₯ + 3y = 14 3/π‘₯ – 4y = 23 4/π‘₯ + 3y = 14 3/π‘₯ – 4y = 23 So, our equations become 4u + 3y = 14 3u – 4y = 23 Now, our equations are 4u + 3y = 14 …(3) 3u – 4y = 23 …(4) From (3) 4u + 3y = 14 4u = 14 – 3y u = (14 βˆ’3𝑦)/4 Putting value of u in (4) 3u – 4y = 23 3 ((14 βˆ’ 3𝑦)/4) – 4y = 23 Multiplying both sides by 4 4 Γ— 3 ((14 βˆ’ 3𝑦)/4) – 4 Γ— 4y = 4 Γ— 23 3(14 – 3y) βˆ’ 16y = 92 42 – 9y βˆ’ 16y = 92 – 9y – 16y = 92 – 42 –25y = 50 y = 50/(βˆ’25) y = – 2 Putting y = – 2 in equation (3) 4u + 3y = 14 4u + 3(–2) = 14 4u – 6 = 14 4u = 14 + 6 u = 20/4 u = 5 But u = 1/π‘₯ 5 = 1/π‘₯ x = 𝟏/πŸ“ Hence, x = 𝟏/πŸ“, y = –2 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: (iv) 5/(π‘₯ βˆ’ 1) + 1/(𝑦 βˆ’ 2) = 2 6/(π‘₯ βˆ’ 1) βˆ’ 3/(𝑦 βˆ’ 2) = 1 5/(π‘₯ βˆ’ 1) + 1/(𝑦 βˆ’ 2) = 2 6/(π‘₯ βˆ’ 1) βˆ’ 3/(𝑦 βˆ’ 2) = 1 So, our equations become 5u + v = 2 6u – 3v = 1 Our equations are 5u + v = 2 …(3) 6u – 3v = 1 …(4) From (3) 5u + v = 2 v = 2 – 5u Putting value of v in (4) 6u – 3v = 1 6u – 3(2 – 5u) = 1 6u – 6 + 15u = 1 6u + 15u = 1 + 6 21u = 7 u = 7/21 u = 𝟏/πŸ‘ Putting u = 1/3 in (3) 5u + v = 2 5 (1/3) + v = 2 5/3 + v = 2 v = 2 – 5/3 v = (2(3) βˆ’ 5)/3 v = 𝟏/πŸ‘ Hence, u = 1/3 & v = 1/3 We need to find x & y We know that u = 𝟏/(𝒙 βˆ’ 𝟏) 1/3 = 1/(π‘₯ βˆ’ 1) x – 1 = 3 x = 3 + 1 x = 4 v = 𝟏/(π’š βˆ’ 𝟐) 1/3 = 1/(𝑦 βˆ’2) y – 2 = 3 y = 3 + 2 y = 5 So, x = 4, y = 5 is the solution of our equations

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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.