Last updated at Dec. 13, 2024 by Teachoo
Question 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 Cost of one ball be Rs y Given that 3 bats and 6 balls cost Rs 3900 3 × Cost of one bat + 6 × Cost of one ball = 3900 3x + 6y = 3900 3(x + 2y) = 3 × 1300 x + 2y = 1300 Also, She buys another bat and 3 more balls of the same kind for Rs 1300 1 × Cost of one bat + 3 × Cost of one ball = 1300 x + 3y = 1300 Now, plotting equations x + 2y = 1300 ...(1) x + 3y = 1300 ...(2) For Equation (1) x + 2y = 1300 Let x = 0 0 + 2y = 1300 2y = 1300 y = 1300/2 y = 650 So, x = 0, y = 650 is a solution i.e. (0, 650) is a solution Let x = 100 100 + 2y = 1300 2y = 1300 − 100 y = 1200/2 y = 600 So, x = 100, y = 600 is a solution i.e. (100, 600) is a solution For Equation (2) x + 3y = 1300 Let x = 0 0 + 3y = 1300 3y = 1300 y = 1300/3 y = 433.3 So, x = 0, y = 433.33.. is a solution i.e. (0, 433.33..) is a solution Let x = 100 100 + 3y = 1300 3y = 1300 − 100 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
Graph of pair of Linear Equations
About the Author
Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo