# Ex 3.1, 2 - Chapter 3 Class 10 Pair of Linear Equations in Two Variables

Last updated at Dec. 28, 2018 by Teachoo

Last updated at Dec. 28, 2018 by Teachoo

Transcript

Ex 3.1,2 The coach of a cricket team buys 3 bats and 6 balls for Rs 3900. Later, she buys another bat and 3 more balls of the same kind for Rs 1300. Represent this situation algebraically and geometrically. Let the cost of one bat be Rs x & Let the cost of one ball be Rs y 3 bats and 6 balls cost Rs 3900 3 × Cost of one bat + 6 × Cost of one ball = 3900 3x + 6y = 3900 3x + 6y – 3900 = 0 3(x + 2y – 1300) = 0 (x + 2y – 1300) = 0/3 x + 2y – 1300 = 0 Also, 1 bat and 3 balls cost Rs 1300 1 × Cost of one bat + 3 × Cost of one ball = 1300 x + 3y = 1300 x + 3y – 1300 = 0 Now, plotting equations x + 2y – 1300 = 0 ...(1) x + 3y – 1300 = 0 Solving equation (1) , x + 2y – 1300 = 0 2y = 1300 - x y = (1300 − 𝑥)/2 Let x = 0, y = (1300 − 0)/2 y = (1300 )/2 y = 650 So, x = 0, y = 650 is a solution ,i.e., (0,650) is a solution Let x = 100 y = (1300 − 100)/2 y = (1200 )/2 y = 600 So, x = 100, y = 600 is a solution ,i.e., (100,600) is a solution Solving equation (2) , x + 3y – 1300 = 0 3y = 1300 - x y = (1300 − 𝑥)/3 Let x = 0, y = (1300 − 0)/3 y = (1300 )/3 y = 433.33 So, x = 0, y = 433.33 is a solution ,i.e., (0,433.33) is a solution. Let x = 100 y = (1300 − 100)/3 y = (1200 )/3 y = 400 So, x = 100, y = 400 is a solution ,i.e., (100,400) is a solution We will plot both equations on the graph

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.