Question

2. The coach of a cricket team buys 3 bats and 6 balls for 3900. Later, she buys another
bat and 3 more balls of the same kind for 1300. Represent this situation algebraically
and geometrically. ​

Answer 1

Answer:

let the bats be x

let the balls be y

1st equation

3x+6y=3900. 2nd equation

x+3y=1300

add equation 1 and 2

=4x+9y = 5200

Answer 2

Answer:

let the cost of 1 bat be x .

cost of 1 ball be y.

case1: cost of 3 bat = 3x

cost of 6 balls = 6y

ATQ

3x + 6y = 3900 ......(1)

case 2: cost of 4 bat = 4x

cost of 9 ball = 9y

ATQ

4x+ 9y = 3900+ 1300

or 4x + 9y = 5200..........(2)

3x + 6y = 3900.......(1) × 3

4x+ 9y = 5200........(2)× 2

subtracting 2 from 1

9x +18y = 10800

8x+ 18y = 10400

- - -

x = 400

putting x in (1)

3(400) + 6y = 3900

1200 + 6y = 3900

6y = 3900-1200

6y = 2700

y = 2700÷ 6

y = 450

cost of 1 bat = 400.

cost of 1ball = 450.