Last updated at June 22, 2017 by Teachoo

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Ex 3.6, 1 Solve the following pairs of equations by reducing them to a pair of linear equations: (iii) 4/๐ฅ + 3y = 14 3/๐ฅ โ 4y = 23 4/๐ฅ + 3y = 14 3/๐ฅ โ 4y = 23 Now, our equations are 4u + 3y = 14 โฆ(3) 3u โ 4y = 23 โฆ(4) From (3) 4u + 3y = 14 4u = 14 โ 3y u = (14 โ3๐ฆ)/4 Putting value of u in (4) 3u โ 4y = 23 3((14 โ3๐ฆ)/4) โ 4y = 23 Multiplying both sides by 4 4 ร 3((14 โ3๐ฆ)/4) โ 4 ร 4y = 4 ร 23 3(14 โ 3y ) - 16y = 92 42 โ 9y - 16y = 92 โ 9y โ 16y = 92 โ 42 โ25y = 50 y = 50/(โ25) = โ 2 Putting y = โ 2 in equation (3) 4u + 3y = 14 4u + 3(โ2) = 14 4u โ 6 = 14 4u = 14 + 6 4u = 20 u = 20/4 u = 5 But u = 1/๐ฅ 5 = 1/๐ฅ x = 1/5 Hence, x = 1/5 , y = โ2 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/(๐ฅ โ1) + 1/(๐ฆ โ2) = 2 6/(๐ฅ โ1) - 3/(๐ฆ โ2) = 1 5/(๐ฅ โ1) + 1/(๐ฆ โ2) = 2 6/(๐ฅ โ1) - 3/(๐ฆ โ2) = 1 Our equations are 5u + v = 2 โฆ(3) 6u โ 3v = 1 โฆ(4) From (3) 5u + v = 2 v = 2 โ 5u Putting value of v in (4) 6u โ 3v = 1 6u โ 3(2 โ 5u) = 1 6u โ 6 + 15u = 1 6u + 15u = 1 + 6 21u = 7 u = 7/21 u = 1/3 Putting u = 1/3 in (3) 5u + v = 2 5(1/3) + v = 2 5/3 + v = 2 v = 2 โ 5/3 v = (2(3) โ 5)/3 v = (6 โ 5)/3 v = 1/3 Hence, u = 1/3 & v = 1/3 Hence, u = 1/3 & v = 1/3 We need to find x & y We know that So, x = 4, y = 5 is the solution of our equations

Example 6
Important

Example 10 Important

Example 11 Important

Example 13 Important

Example 14 Important

Example 19 Important

Ex 3.2, 1 Important

Ex 3.2, 3 Important

Ex 3.2, 6 Important

Ex 3.2, 7 Important

Ex 3.5, 2 Important

Ex 3.6, 1 (i) and (ii) Important

Ex 3.6, 1 (iii) and (iv) Important You are here

Ex 3.6, 1 (v) and (vi) Important

Ex 3.6, 1 (vii) and (viii) Important

Ex 3.6, 2 Important

Class 10

Important Questions for Exam - Class 10

- Chapter 1 Class 10 Real Numbers
- Chapter 2 Class 10 Polynomials
- Chapter 3 Class 10 Pair of Linear Equations in Two Variables
- Chapter 4 Class 10 Quadratic Equations
- Chapter 5 Class 10 Arithmetic Progressions
- Chapter 6 Class 10 Triangles
- Chapter 7 Class 10 Coordinate Geometry
- Chapter 8 Class 10 Introduction to Trignometry
- Chapter 9 Class 10 Some Applications of Trignometry
- Chapter 10 Class 10 Circles
- Chapter 11 Class 10 Constructions
- Chapter 12 Class 10 Areas related to Circles
- Chapter 13 Class 10 Surface Areas and Volumes
- Chapter 14 Class 10 Statistics
- Chapter 15 Class 10 Probability

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.