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IceJOKER [234]
2 years ago
12

Someone can help me for all the question ?

Mathematics
1 answer:
Dvinal [7]2 years ago
4 0

Answer:

1. 5

2. 3 adult tickets; 2 child tickets

3. See below

Step-by-step explanation:

1.

2 adult tickets cost 2 × $12 = $24

Subtract $24 from $64:

$64 - $24 = $40

Divide $40 by $8:

$40 ÷ $8 = 5

Answer: 5 child tickets

2.

x = number of adult tickets

y = number of child tickets

The total number of tickets is 5, so the first equation is:

x + y = 5

The cost of x adult tickets is 12x. The cost of y child tickets is 8y. The total cost is 12x + 8y. We are told the total cost is $52. The second equation is:

12x + 8y = 52

The system of equation is:

x + y = 5

12x + 8y = 52

Let's solve it by substitution.

Solve the first equation for x.

x = 5 - y

Now we substitute x in the second equation by 5 - y.

12(5 - y) + 8y = 52

Distribute:

60 - 12y + 8y = 52

Combine like terms on the left side:

-4y + 60 = 52

Subtract 60 from both sides:

-4y = -8

Divide both sides by 4.

y = 2

Substitute y = 2 in the first original equation and solve for x.

x + y = 5

x + 2 = 5

x = 3

Answer: 3 adult tickets; 2 child tickets

3.

The customer's claim is incorrect. His sum is $71. 71 is an odd number. Since all ticket prices are even numbers, it is impossible to add only even numbers and get an odd sum. A sum of several amounts of 12 and several amounts of 8 can never equal 71.

We can also show his claim is false using a system of equations.

x + y = 7

12x + 8y = 71

Use substitution.

x = 7 - y

12(7 - y) + 8y = 71

84 - 12y + 8y = 71

84 - 4y = 71

-4y = -13

y = 3.25

This means 3.25 child tickets. The problem states that tickets can be purchased only in whole numbers, so 3.25 tickets cannot be purchased, so the customer's claim is false.

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