Question

The town glee club sold tickets for the summer concert. The club charged $8 for a child ticket and $18 for an adult ticket. A total of $8,400 in ticket sales was raised. The number of adult tickets, a, is 50 more than twice the number of child tickets, c. Which system of equations will solve for the number of each type of ticket?
8 a + 18 c = 8,400. a = 2 c minus 50.
8 c + 18 a = 8,400. a = 2 c + 50.
18 a + 8 c = 8,400. c = 2 a + 50.
8 a minus 18 c = 8,400. a = 2 c minus 50.

Answer 1

To set up the system of equations based on the problem description, we need two key pieces of information:

1. The total revenue from the ticket sales.
2. The relationship between the number of adult tickets and child tickets.

From the problem:

1. The revenue equation is based on ticket prices:
- The ticket prices are $8 for child tickets (c) and $18 for adult tickets (a). Thus, the total revenue equation can be expressed as:
\[
8c + 18a = 8400
\]

2. The relationship between adult tickets and child tickets:
- The problem states that the number of adult tickets, \( a \), is 50 more than twice the number of child tickets, \( c \). This can be written as:
\[
a = 2c + 50
\]

Putting these together, we have the following system of equations:
\[
\begin{align*}
1) & \quad 8c + 18a = 8400 \\
2) & \quad a = 2c + 50
\end{align*}
\]

Thus, the correct answer is:
**8 c + 18 a = 8,400. a = 2 c + 50.**