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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 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.