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From the image, we have a system of linear equations problem related to a barbecue dinner fundraiser, which involves finding the number of child and adult plates sold. Here's a breakdown of the problem:

- The total number of dinners sold: **464**.
- The total revenue collected: **$2,455.90**.
- The cost per child plate: **$3.20**.
- The cost per adult plate: **$7.10**.

We can set up the following system of equations:

1. Let $ c $ be the number of child plates and $ a $ be the number of adult plates.
   
   The total number of dinners sold gives the equation:
   \[
   c + a = 464
   \]

2. The total revenue collected gives the second equation:
   \[
   3.20c + 7.10a = 2455.90
   \]

### Steps to solve:
We can solve this system of equations using substitution or elimination.

1. From the first equation, solve for $ c $:
   \[
   c = 464 - a
   \]

2. Substitute $ c = 464 - a $ into the second equation:
   \[
   3.20(464 - a) + 7.10a = 2455.90
   \]

3. Distribute the $ 3.20 $ on the left-hand side:
   \[
   3.20 \times 464 - 3.20a + 7.10a = 2455.90
   \]
   \[
   1484.80 - 3.20a + 7.10a = 2455.90
   \]

4. Combine like terms:
   \[
   1484.80 + 3.90a = 2455.90
   \]

5. Subtract $ 1484.80 $ from both sides:
   \[
   3.90a = 971.10
   \]

6. Solve for $ a $:
   \[
   a = \frac{971.10}{3.90} \approx 249
   \]

7. Now, substitute $ a = 249 $ back into the first equation to find $ c $:
   \[
   c + 249 = 464
   \]
   \[
   c = 464 - 249 = 215
   \]

### Conclusion:
- The number of child plates sold is **215**.
- The number of adult plates sold is **249**.

Would you like a detailed breakdown of the steps, or do you have any other questions?

### Related Questions:
1. How would the system change if the price of the child plate increased to $4.00?
2. What would happen if only 400 dinners were sold, but the same amount of money was collected?
3. How would you solve this system using the elimination method instead of substitution?
4. Can you verify the solution by substituting the values of $ c $ and $ a $ back into both original equations?
5. What if the total revenue had a small error and was recorded as $2,460? How would that affect the solution?

### Tip:
When solving systems of equations, substitution is useful when one equation is easy to solve for one variable, while elimination works well when the coefficients of one variable are already aligned.

The Downtown Community Barbecue served 464 dinners. A child’s plate cost $3.20 and an adult’s plate cost $7.10. A total of $2,455.90 was collected. How many of each type of plate was served?

Learn how to solve a system of linear equations to determine the number of child and adult plates sold at a barbecue dinner. The problem involves key algebraic concepts and uses substitution to find the solution. Suitable for students in Grades 9-11.

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