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 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

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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