- Identify the relationship between the variables: the total number of tickets is the sum of student and adult tickets, \(c + a = 67\).
- Isolate the variable \(c\) in terms of \(a\) by subtracting \(a\) from both sides: \(c = 67 - a\).
- Compare this expression with the provided options.
- The correct expression is \(\boxed{67 - a}\).
Explanation
- Problem Analysis
We are given that a school production sold student tickets for \(\$5\) and adult tickets for \(\$10\). We are told that a total of \(67\) tickets were sold for a total revenue of \(\$440\). Let \(c\) represent the number of student tickets sold and \(a\) represent the number of adult tickets sold. We can set up the following equations based on the information provided:
- Total number of tickets: \(c + a = 67\)
- Total revenue: \(5c + 10a = 440\)
The table asks for an expression that represents \(c\), the number of student tickets sold.
- Solving for c
We need to express the number of student tickets, \(c\), in terms of the number of adult tickets, \(a\). Starting with the total ticket count equation:
\[c + a = 67\]
To isolate \(c\), we subtract \(a\) from both sides of the equation:
\[c = 67 - a\]
This shows that the number of student tickets is equal to the total number of tickets sold minus the number of adult tickets sold. Comparing this to the given options, \(67 - a\) is the correct expression.
- Final Conclusion
The expression that replaces \(c\) in the table is \(\boxed{67 - a}\).
Examples
This problem shows how to translate real-world constraints into algebraic equations. For example, if you are organizing a bake sale where cookies cost \(\$2\) and brownies cost \(\$3\), and you know you sold 50 total items for \(\$120\), you can use the same method of defining variables and setting up a system of equations to determine exactly how many of each item were sold. This skill is foundational for managing budgets, inventory, and resource allocation in everyday tasks.