A movie theater has a seating capacity of 227. The theater charges $5.00 for children, $7.00 for students, and $12.00 of adults. There are half as many adults as there are children. If the total ticket sales was $ 1640, How many children, students, and adults attended?


This problem involves solving a system of equations. Let's define the variables and set up the equations based on the given information.

### Step 1: Define the Variables
- Let $ c $ represent the number of children.
- Let $ s $ represent the number of students.
- Let $ a $ represent the number of adults.

### Step 2: Set Up the Equations

1. **Equation for seating capacity**:
   The total number of people (children + students + adults) is equal to the seating capacity:
   \[
   c + s + a = 227
   \]

2. **Equation for ticket sales**:
   The total revenue from ticket sales is $1640:
   \[
   5c + 7s + 12a = 1640
   \]

3. **Relationship between adults and children**:
   The problem states that there are half as many adults as there are children:
   \[
   a = \frac{1}{2}c
   \]

### Step 3: Solve the System of Equations

#### Substitute $ a = \frac{1}{2}c $ into the other equations:
1. **Substitute into the seating capacity equation**:
   \[
   c + s + \frac{1}{2}c = 227
   \]
   Combine like terms:
   \[
   \frac{3}{2}c + s = 227
   \]
   Multiply the entire equation by 2 to eliminate the fraction:
   \[
   3c + 2s = 454 \quad \text{(Equation 1)}
   \]

2. **Substitute into the ticket sales equation**:
   \[
   5c + 7s + 12\left(\frac{1}{2}c\right) = 1640
   \]
   Simplify the equation:
   \[
   5c + 7s + 6c = 1640
   \]
   Combine like terms:
   \[
   11c + 7s = 1640 \quad \text{(Equation 2)}
   \]

### Step 4: Solve the System of Two Equations

We now have two equations:

1. $ 3c + 2s = 454 $
2. $ 11c + 7s = 1640 $

We can solve this system using the substitution or elimination method.

#### Multiply the first equation by 7 and the second by 2 to eliminate $ s $:

\[
7(3c + 2s) = 7 \times 454 \implies 21c + 14s = 3178 \quad \text{(Equation 3)}
\]

\[
2(11c + 7s) = 2 \times 1640 \implies 22c + 14s = 3280 \quad \text{(Equation 4)}
\]

#### Subtract Equation 3 from Equation 4 to solve for $ c $:

\[
22c + 14s - (21c + 14s) = 3280 - 3178
\]
\[
c = 102
\]

Now that we have $ c = 102 $, substitute it back into Equation 1:

\[
3(102) + 2s = 454
\]
\[
306 + 2s = 454
\]
\[
2s = 148
\]
\[
s = 74
\]

Finally, substitute $ c = 102 $ into $ a = \frac{1}{2}c $ to find $ a $:

\[
a = \frac{1}{2} \times 102 = 51
\]

### Final Answer
- Number of children $ c = 102 $
- Number of students $ s = 74 $
- Number of adults $ a = 51 $

Would you like further details or have any questions?

Here are 5 related questions to expand your understanding:

1. How would the solution change if the seating capacity was different?
2. What if the total ticket sales were $2000? How would you approach the problem?
3. How can you verify that the solution satisfies all the conditions of the problem?
4. If the theater decided to increase the ticket prices, how would that affect the solution?
5. What would the equation look like if there were twice as many adults as children?

**Tip:** In word problems, always double-check the relationships described (e.g., "half as many," "total sales," etc.) to ensure you've correctly set up your equations.

Learn how to solve a word problem involving ticket sales and seating capacity at a movie theater. This problem requires setting up and solving a system of equations to find the number of children, students, and adults attending the theater. Suitable for high school students.

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