Let's denote the price of a student ticket by $x$ dollars and the price of an adult ticket by $y$ dollars.

We are given two equations based on the ticket sales:

1. On the first day: $200x + 25y = 2775$
2. On the second day: $247x + 42y = 3594$

We can solve this system of linear equations to find the values of $x$ and $y$.

Step 1: Solve the first equation for $y$

$200x + 25y = 2775$ $25y = 2775 - 200x$ $y = \frac{2775 - 200x}{25}$ $y = 111 - 8x$

Step 2: Substitute $y = 111 - 8x$ into the second equation

$247x + 42(111 - 8x) = 3594$ $247x + 4662 - 336x = 3594$ $-89x + 4662 = 3594$ $-89x = 3594 - 4662$ $-89x = -1068$ $x = \frac{-1068}{-89} = 12$

Step 3: Substitute $x = 12$ back into the equation for $y$

$y = 111 - 8(12)$ $y = 111 - 96$ $y = 15$

Final Answer:

• The price of a student ticket ($x$) is $12.
• The price of an adult ticket ($y$) is $15.

Would you like further details or have any questions?

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Related Questions:

1. How would the system of equations change if the school sold different numbers of tickets?
2. What if the total earnings from the first day were $3000 instead?
3. How could you solve these equations graphically?
4. What happens if both tickets were priced the same?
5. How would you solve this problem using matrix methods?

Tip: When solving linear equations, isolating one variable first can simplify the process significantly.