Let the number of adults be \( a \) and the number of children be \( c \).
We have two equations based on the problem:
-
The total number of attendees (adults and children) is 500: \[ a + c = 500 \]
-
The total revenue from admissions should be $3,500. Adults are charged $10 and children $5: \[ 10a + 5c = 3500 \]
Now, we have a system of equations:
- \( a + c = 500 \) (1)
- \( 10a + 5c = 3500 \) (2)
We can simplify equation (2) by dividing all terms by 5: \[ 2a + c = 700 \]
Now we have the simplified system:
- \( a + c = 500 \) (1)
- \( 2a + c = 700 \) (3)
Next, we can eliminate \( c \) by subtracting equation (1) from equation (3): \[ (2a + c) - (a + c) = 700 - 500 \] \[ 2a + c - a - c = 200 \] \[ a = 200 \]
Now that we have \( a = 200 \), we can substitute it back into equation (1) to find \( c \): \[ 200 + c = 500 \] \[ c = 500 - 200 \] \[ c = 300 \]
Thus, the number of adults and children that need to attend is:
- Adults: \( 200 \)
- Children: \( 300 \)
To verify, we can check the revenue: \[ 10a + 5c = 10(200) + 5(300) = 2000 + 1500 = 3500 \] And the total attendance: \[ a + c = 200 + 300 = 500 \]
Therefore, the solution is correct. The Kesling Middle School PTA needs 200 adults and 300 children to attend to reach their goal of $3,500.