Last updated at May 29, 2018 by Teachoo

Transcript

Ex 3.6 , 1 Solve the following pairs of equations by reducing them to a pair of linear equations: (i) 1/2๐ฅ + 1/3๐ฆ = 2 1/3๐ฅ + 1/2๐ฆ = 13/6 1/2๐ฅ + 1/3๐ฆ = 2 1/3๐ฅ + 1/2๐ฆ = 13/6 So, our equations become Our equations are 3u + 2v = 12 โฆ(3) 2u + 3v = 13 โฆ(4) From (3) 3u + 2v = 12 3u = 12 โ 2v u = (12 โ2๐ฃ)/3 Putting value of u in (4) 2u + 3v = 13 2((12 โ2๐ฃ)/3) + 3v = 13 Multiplying both sides by 3 3 ร 2((12 โ2๐ฃ)/3) + 3 ร 3v = 3 ร 13 2(12 โ 2v) + 9v = 39 24 โ 4v + 9v = 39 โ 4v + 9v = 39 โ 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 โ 6 3u = 6 u = 6/3 u = 2 Hence v = 3, u = 2 But we have to find x & y We know that So, x = 1/2 , y = 1/3 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/โ๐ฅ + 3/โ๐ฆ = 2 4/โ๐ฅ โ 9/โ๐ฆ = โ1 2/โ๐ฅ + 3/โ๐ฆ = 2 4/โ๐ฅ โ 9/โ๐ฆ = -1 Our equations 2u + 3v = 2 โฆ(3) 4u โ 9v = โ1 โฆ(4) From (3) 2u + 3v = 2 2u = 2 โ 3v u = (2 โ3๐ฃ)/2 Putting value of u in (4) 4u โ 9v = โ 1 4((2 โ3๐ฃ)/2) โ 9v = โ 1 2(2 โ 3v) โ 9v = โ 1 4 โ 6v โ 9v = โ 1 โ 6v โ 9v = โ 1 โ 4 โ 15v = โ 5 v = (โ5)/(โ15) v = 1/3 Putting v = 1/3 in (3) 2u + 3v = 2 2u + 3(1/3) = 2 2u + 1 = 2 2u = 2 โ 1 2u = 1 u = 1/2 Hence u = 1/2 & v = 1/3 But, we need to find x & y We need to find x & y Therefore, x = 4, y = 9 is the solution of the given equation

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 You are here

Ex 3.6, 1 (iii) and (iv) Important

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.