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Question
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.
Solution
Let cost of one bat = Rs x
Cost of one ball = Rs y
3 bats and 6 balls for Rs 3900 So that
3x + 6y = 3900
Dividing equation by 3, we get
x + 3y = 1300
Graphical representation:
we have,
3x + 6y = 3900
3(x + 2y) = 3900
x + 2y = 1300
a = 1300 - 2y
Thus, we have following table:
| x | 1300 | 500 | 700 | 100 |
| y | 0 | 400 | 300 | 600 |
we have, x + 3y = 1300
x = 1300 - 3y
Thus, we have following table:
| x | 1300 | 100 | 400 | 1000 |
| y | 0 | 400 | 300 | 100 |
when we plot the graph of equations, we find that both the lines intersect at the point (1300, 0).
Therefore, a = 1300, y = 0 is the solution of the given system of equations.
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