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Answer :
Final answer:
The price of one senior ticket is $8, and the price of one child ticket is $3. This conclusion is reached by setting up and solving a system of linear equations derived from the two days of ticket sales data provided.
Explanation:
To find the prices of one senior ticket and one child ticket from the information given, we need to set up a system of linear equations based on the two days of ticket sales.
Let x represent the price of a senior ticket and y represent the price of a child ticket.
From the first day, we have the equation 45x + 15y = 405 since 45 senior tickets and 15 child tickets were sold for a total of $405.
From the second day, we have the equation 52x + 15y = 461 since 52 senior tickets and 15 child tickets were sold for a total of $461.
To solve this system, first multiply the first equation by -1 to help eliminate y:
- -1(45x + 15y) = -1(405) leads to -45x - 15y = -405
- Add this to the second equation: (-45x - 15y) + (52x + 15y) = (-405) + 461 which simplifies to 7x = 56
- Divide both sides by 7 to find x: x = 56 / 7, which gives x = 8
Now that we have the price of a senior ticket, we can substitute x = 8 back into the first equation (45x + 15y = 405):
- 45(8) + 15y = 405
- 360 + 15y = 405
- 15y = 405 - 360
- 15y = 45
- y = 45 / 15
- y = 3
Therefore, the price of one senior ticket is $8 and the price of one child ticket is $3.
Learn more about System of Linear Equations here:
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