Learn all Concepts of Chapter 3 Class 10 (with VIDEOS). Check - Linear Equations in 2 Variables - Class 10

Last updated at May 2, 2020 by Teachoo

Learn all Concepts of Chapter 3 Class 10 (with VIDEOS). Check - Linear Equations in 2 Variables - Class 10

Transcript

Ex 3.6, 1 Solve the following pairs of equations by reducing them to a pair of linear equations: (v) (7๐ฅ โ 2๐ฆ)/๐ฅ๐ฆ = 5 (8๐ฅ + 7๐ฆ)/๐ฅ๐ฆ = 15 Given (7๐ฅ โ 2๐ฆ)/๐ฅ๐ฆ = 5 (7๐ฅ )/๐ฅ๐ฆ โ (2๐ฆ )/๐ฅ๐ฆ = 5 (7 )/๐ฆ โ(2 )/๐ฅ = 5 (โ๐ )/๐ +(๐ )/๐ = 5 (8๐ฅ + 7๐ฆ)/๐ฅ๐ฆ = 15 (8๐ฅ )/๐ฅ๐ฆ + (7๐ฆ )/๐ฅ๐ฆ = 15 (8 )/๐ฆ +(7 )/๐ฅ = 15 (๐ )/๐ +(๐ )/๐ = 15 Our equations are (โ2 )/๐ฅ +(7 )/๐ฆ = 5 โฆ(1) (7 )/๐ฅ +(8 )/๐ฆ = 15 ...(2) Let 1/๐ฅ = u & 1/๐ฆ = v So, our equations become โ2u + 7v = 5 7u + 8v = 15 Hence, we solve โ2u + 7v = 5 โฆ(3) 7u + 8v = 15 โฆ(4) From (3) โ2u + 7v = 5 7v = 5 + 2u v = (5 + 2๐ข)/7 Putting value of v in (4) 7u + 8v = 15 7u + 8((5 + 2๐ข)/7) = 15 Multiplying 7 both sides 7 ร 7u + 7 ร 8 ((5 + 2๐ข)/7) = 7 ร 15 49u + 8(5 + 2u) = 105 49u + 40 + 16u = 105 49u + 16u = 105 โ 40 65u = 65 u = 65/65 u = 1 Putting value of u in (3) โ2u + 7v = 5 โ2(1) + 7v = 5 โ2 + 7v = 5 7v = 5 + 2 7v = 7 v = 7/7 v = 1 Hence, u = 1, v = 1 But we have to find x & y We know that u = ๐/๐ 1 = 1/๐ฅ x = 1 v = ๐/๐ 1 = 1/๐ฆ y = 1 Hence, x = 1 , y = 1 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: (vi) 6x + 3y = 6xy 2x + 4y = 5xy Given 6x + 3y = 6xy Diving whole equation by xy (6๐ฅ + 3๐ฆ)/๐ฅ๐ฆ = 6๐ฅ๐ฆ/๐ฅ๐ฆ 6๐ฅ/๐ฅ๐ฆ +3๐ฆ/๐ฅ๐ฆ = 6 ๐/๐ +๐/๐ = 6 2x + 4y = 5xy Diving whole equation by xy (2๐ฅ + 4๐ฆ)/๐ฅ๐ฆ = 5๐ฅ๐ฆ/๐ฅ๐ฆ 2๐ฅ/๐ฅ๐ฆ +4๐ฆ/๐ฅ๐ฆ = 5 ๐/๐ +๐/๐ = 5 Hence, our equations are 6/๐ฆ +3/๐ฅ = 6 โฆ(1) 2/๐ฆ +4/๐ฅ = 5 โฆ(2) Let 1/๐ฅ = u & 1/๐ฆ = v So, our equations become 6v + 3u = 6 2v + 4u = 5 Now, we solve 6v + 3u = 6 โฆ(3) 2v + 4u = 5 โฆ(4) From (3) 6v + 3u = 6 6v = 6 โ 3u v = (6 โ 3๐ข)/6 Putting value of v in (4) 2v + 4u = 5 2((6 โ 3๐ข)/6) + 4u = 5 ((6 โ 3๐ข)/3) + 4u = 5 Multiplying both sides by 3 3 ร ((6 โ3๐ข)/3) + 3 ร 4u = 3 ร 5 (6 โ 3u) + 12u = 15 โ3u + 12u = 15 โ 6 9u = 9 u = 9/9 u = 1 Putting u = 1 in (3) 6v + 3u = 6 6v + 3(1) = 6 6v + 3 = 6 6v = 6 โ 3 6v = 3 v = 3/6 v = ๐/๐ Hence, u = 1 , v = 1/2 But we have to find x & y Now, u = ๐/๐ 1 = 1/๐ฅ x = 1 v = ๐/๐ 1/2 = 1/๐ฆ y = 2 Hence, x = 1 , y = 2 is the solution of the given equation

Chapter 3 Class 10 Pair of Linear Equations in Two Variables

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 3.5, 2 Important

Ex 3.2, 7 Important

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

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

Ex 3.6, 1 (v) and (vi) Important You are here

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

Ex 3.6, 2 (i) 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 9 years. He provides courses for Maths and Science at Teachoo.