Cindy is selling circus tickets. On the first day she sells 6 adult tickets and 5 children tickets for total of 112.50. On the second day she sells 8 adult tickets and 4 children tickets for the total 130. Determine the cost of adult and child tickets. Used a to represent adult and c to represent child tickets. Write an equation to represent the sales of day 1 and day 2. Used the elimination method.

Using the concept of elimination, the cost of adult and children tickets are $12.5 and $7.5 respectively

Let :

Creating a system of equation thus :

6a + 5c = 112.50 - - - (1)

8a + 4c = 130 - - - - (2)

Multiply (1) by 4 and (2) by 5

24a + 20c = 450 - - - - (3)

40a + 20c = 650 - - - - (4)

Subtract

-16a = - 200

Divide both sides by - 16

a = 12.5

From (2) :

8(12.5) + 4c = 130

100 + 4c = 130

4c = 130 - 100

4c = 30

c = 7.5

Hence, the cost of adult and children tickets are $12.5 and $7.5

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Children tickets cost 7.50

Adult tickets cost 12.50

Step-by-step explanation:

Let a represent adult tickets

Let c represent child tickets

On the first day she sells 6 adult tickets and 5 children tickets for total of 112.50. This can be written as:

6a + 5c = 112.50 equation i

On the second day she sells 8 adult tickets and 4 children tickets for the total 130. This can be written as:

8a + 4c = 130 equation ii

6a + 5c = 112.50 equation i

8a + 4c = 130 equation ii

Multiply equation i by 4

Multiply equation ii by 5

24a + 20c = 450 equation iii

40a + 20c = 650 equation iv

Subtract iii from iv

16a = 200

a = 200/16

a = 12.50

Adult tickets cost 12.50

From equation ii,

8a + 4c = 130

8(12.50) + 4c = 130

100 + 4c = 130

4c = 130 - 100

4c = 30

c = 30/4

c = 7.50

Children tickets cost 7.50

substitute,

circle 1 = 5

circle 2 = 20

circle 3 = 44

in this case no matter how you add, you will always get a perfect square..

hence proved

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