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Let's solve this problem step-by-step using algebra.

### Step 1: Define the variables
Let:
- $ C $ be the number of children.
- $ S $ be the number of students.
- $ A $ be the number of adults.

### Step 2: Write the equations based on the problem conditions

1. **Total number of attendees equation:**
   \[
   C + S + A = 303
   \]

2. **Relationship between the number of children and adults:**
   \[
   A = \frac{1}{2}C
   \]

3. **Total ticket sales equation:**
   \[
   5C + 7S + 12A = 2208
   \]

### Step 3: Substitute the relationship $ A = \frac{1}{2}C $ into the equations

Substituting $ A = \frac{1}{2}C $ into the total number of attendees equation:
\[
C + S + \frac{1}{2}C = 303
\]
Multiplying the entire equation by 2 to clear the fraction:
\[
2C + 2S + C = 606
\]
\[
3C + 2S = 606 \quad \text{(Equation 1)}
\]

Substituting $ A = \frac{1}{2}C $ into the ticket sales equation:
\[
5C + 7S + 12\left(\frac{1}{2}C\right) = 2208
\]
\[
5C + 7S + 6C = 2208
\]
\[
11C + 7S = 2208 \quad \text{(Equation 2)}
\]

### Step 4: Solve the system of equations

We now have the following system of equations:
1. $ 3C + 2S = 606 $
2. $ 11C + 7S = 2208 $

Let's solve these equations using elimination or substitution. I'll proceed with elimination for clarity.

Multiply Equation 1 by 7 to align with the second equation:
\[
21C + 14S = 4242
\]

Multiply Equation 2 by 2 to align with the first equation:
\[
22C + 14S = 4416
\]

Subtract the first modified equation from the second:
\[
(22C + 14S) - (21C + 14S) = 4416 - 4242
\]
\[
C = 174
\]

### Step 5: Substitute $ C = 174 $ back into Equation 1

\[
3(174) + 2S = 606
\]
\[
522 + 2S = 606
\]
\[
2S = 84
\]
\[
S = 42
\]

### Step 6: Find the number of adults using $ A = \frac{1}{2}C $

\[
A = \frac{1}{2}(174) = 87
\]

### Final Answer
- **Number of children (C):** 174
- **Number of students (S):** 42
- **Number of adults (A):** 87

Would you like more details on any of these steps, or do you have any questions?

Here are some related questions to expand on this problem:

1. What would be the new total ticket sales if the price for adults was increased to $15?
2. How would the distribution change if the total seating capacity increased to 350?
3. What happens to the number of students if the total number of adults becomes one-third of the children?
4. If the theater gave a discount of $2 for each category, how much would the total ticket sales be?
5. Can this problem be solved using matrix methods or a different algebraic approach?

**Tip:** In word problems, clearly defining variables and writing equations step-by-step makes it easier to organize information and find solutions.

A movie theater has a seating capacity of 303. The theater charges $5.00 for children, $7.00 for students, and $12.00 for adults. There are half as many adults as there are children. If the total ticket sales was $2208, how many children, students, and adults attended?

Learn how to solve a system of linear equations to determine the number of children, students, and adults attending a movie. The problem involves algebraic concepts like solving simultaneous equations, with ticket prices and total sales. Ideal for students in Grades 8-10.

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