st marks community barbecue served 250 dinners a childs plate cost 3.50 and an adukts plate cost 7.00 a total cost of 1347.50was collected how many of each type of plate were served

Let:
- $ x $ represent the number of children's plates served.
- $ y $ represent the number of adult plates served.

We can set up two equations based on the information provided:

1. **Total Plates Served**:
   \[
   x + y = 250
   \]

2. **Total Money Collected**:
   \[
   3.50x + 7.00y = 1347.50
   \]

### Step 1: Simplify the Equations
From the first equation:
\[
y = 250 - x
\]

### Step 2: Substitute $ y $ in the Second Equation
Substitute $ y = 250 - x $ into the second equation:
\[
3.50x + 7.00(250 - x) = 1347.50
\]

Expanding and simplifying:
\[
3.50x + 1750 - 7.00x = 1347.50
\]
\[
-3.50x + 1750 = 1347.50
\]
\[
-3.50x = 1347.50 - 1750
\]
\[
-3.50x = -402.50
\]
\[
x = \frac{-402.50}{-3.50}
\]
\[
x = 115
\]

### Step 3: Substitute $ x $ Back to Find $ y $
Using $ y = 250 - x $:
\[
y = 250 - 115
\]
\[
y = 135
\]

### Solution
- **Children's plates**: 115
- **Adult plates**: 135

Would you like further clarification or have any questions?

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Here are 5 related questions to expand on this solution:

1. How would the equations change if the total amount collected was different?
2. What if the adult plate cost was changed? How would it affect the number of plates?
3. How can this be set up as a system of equations and solved graphically?
4. What if the barbecue served only children's plates? How many would they need to reach a similar total?
5. How can this method be applied to similar cost-based distribution problems?

**Tip:** Always check your solutions by substituting back into the original equations to verify accuracy.

St. Mark's community barbecue served 250 dinners. A child's plate cost $3.50 and an adult's plate cost $7.00. A total of $1347.50 was collected. How many of each type of plate were served?

This solution shows how to calculate the number of children's and adults' plates served at St. Mark's community barbecue, where 250 dinners were served, and $1347.50 was collected. Using systems of equations, this problem covers key algebraic concepts and is ideal for Grades 8-10.

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