Learn all Concepts of Chapter 3 Class 10 (with VIDEOS). Check - Linear Equations in 2 Variables - Class 10     1. Chapter 3 Class 10 Pair of Linear Equations in Two Variables
2. Serial order wise
3. Ex 3.1

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

Ex 3.1 