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 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 … (i)
Dividing equation by 3, we get
x + 2y = 1300
Subtracting 2y both side we get
x = 1300 – 2y
Putting y = -1300, 0 and 1300 we get
x = 1300 – 2 (-1300) = 1300 + 2600 = 3900
x = 1300 -2(0) = 1300 - 0 = 1300
x = 1300 – 2(1300) = 1300 – 2600 = - 1300
| x | 3900 | 1300 | -1300 |
| y | -1300 | 0 | 1300 |
Given that she buys another bat and 2 more balls of the same kind for Rs 1300
So, we get
x + 2y = 1300 … (ii)
Subtracting 2y both side we get
x = 1300 – 2y
Putting y = - 1300, 0 and 1300 we get
x = 1300 – 2 (-1300) = 1300 + 2600 = 3900
x = 1300 – 2 (0) = 1300 - 0 = 1300
x = 1300 – 2(1300) = 1300 – 2600 = -1300
| x | 3900 | 1300 | -1300 |
| y | -1300 | 0 | 1300 |
Algebraic representation
3x + 6y = 3900 … (i)
x + 2y = 1300 … (ii)
Graphical representation,
