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Answer :
Answer:
The price of 1 senior citizen ticket is $4
The price of 1 adult ticket is $7
The price of 1 child ticket is $5
Step-by-step explanation:
Assume that the costs of a senior ticket is $x , an adult ticket is $y and
a child ticket is $z
First day:
The school sold 4 senior citizen tickets, 2 adult tickets and 5 child
tickets for a total of $55
∴ 4x + 2y + 5z = 55 ⇒ (1)
Second day:
The school sold 7 senior citizen tickets, 2 adult tickets and 5 child
tickets for $67
∴ 7x + 2y + 5z = 67 ⇒ (2)
Third day:
The school sold 2 senior citizen tickets, 4 adult tickets and 2 child
tickets for $46
∴ 2x + 4y + 2z = 46 ⇒ (3)
The number of adult tickets and the number of child tickets in the first
and second days are equal, then we can subtract equation (1) from
equation (2) to find x
Subtract equation (1) from equation (2)
∴ (7x - 4x) + (2y - 2y) + (5z - 5z) = 67 - 55
∴ 3x = 12
Divide both sides by 3
∴ x = 4
Substitute the value of x in equation (2)
∴ 7(4) + 2y + 5z = 67
∴ 28 + 2y + 5z = 67
Subtract 28 from both sides
∴ 2y + 5z = 39 ⇒ (4)
Substitute the value of x in equation (3)
∴ 2(4) + 4y + 2z = 46
∴ 8 + 4y + 2z = 46
Subtract 8 from both sides
∴ 4y + 2z = 38 ⇒ (5)
Now lets solve equations (4) and (5) to find y and z
Multiply equation (4) by -2 to eliminate y
∴ -4y - 10z = -78 ⇒ (6)
Add equations (5) and (6)
∴ -8z = -40
Divide both sides by -8
∴ z = 5
Substitute the value of z in equation (4) or (5)
∴ 2y + 5(5) = 39
∴ 2y + 25 = 39
Subtract 25 from both sides
∴ 2y = 14
Divide both sides by 2
∴ y = 7
The price of 1 senior citizen ticket is $4
The price of 1 adult ticket is $7
The price of 1 child ticket is $5
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Final answer:
The problem describes a system of linear equations. By denoting the cost of senior, adult, and child tickets as S, A, and C, we can set up three equations according to the given information. These equations can then be solved using substitution or elimination method.
Explanation:
This problem can be solved using a system of linear equations. Let's denote the cost of one senior citizen ticket as S, one adult ticket as A, and one child ticket as C.
We get these three equations from the problem:
- 4S + 2A + 5C = 55 (from the first day of ticket sales)
- 7S + 2A + 5C = 67 (from the second day of ticket sales)
- 2S + 4A + 2C = 46 (from the third day of ticket sales)
We can solve this system of equations using substitution or elimination method to find the individual prices of the tickets.
Learn more about Linear Equations here:
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