Question 1. The coach of a cricket team buys 3 bats and 6 balls for Rs 3900. Later, she buys another bat and 2 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
3 x + 6y = 3900 … (1)
Divide complete equation by 3 we get
X+ 2y = 1300
Subtract 2y both side we get
X = 1300 – 2y
Plug 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 + 2 y = 1300 … (2)
Subtract 2y both side we get
X = 1300 – 2y
Plug 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
3 x + 6 y = 3900 … (1)
x + 2 y = 1300 … (2)