A ceramies workshop makes serving bowls, platters, and bread baskets to sell at its Winter Festival. A serving bowl takes 3 hours to prepare, 2 hours to paint, and 9 hours to fire. A platter takes 14 hours to prepare, 3 hours to paint, and 4 hours to fire. A bread basket takes 4 hours to prepare, 14 hours to paint, and 7 hours to fire. If the workshop has 92 hours for prep time, 70 hours for painting, and 109 hours for firing, how many of each can be made?

This problem can be solved by setting up a system of linear equations based on the hours required for each activity (preparing, painting, and firing) and the total hours available for each.

Let:
- \( x \) = number of serving bowls
- \( y \) = number of platters
- \( z \) = number of bread baskets

We can now express the constraints based on the available hours for preparation, painting, and firing.

### Preparation time:
- A serving bowl takes 3 hours to prepare, so \( 3x \) hours total for \( x \) serving bowls.
- A platter takes 14 hours to prepare, so \( 14y \) hours total for \( y \) platters.
- A bread basket takes 4 hours to prepare, so \( 4z \) hours total for \( z \) bread baskets.
- The workshop has 92 hours available for preparation.
\[
3x + 14y + 4z = 92
\]

### Painting time:
- A serving bowl takes 2 hours to paint, so \( 2x \) hours total for \( x \) serving bowls.
- A platter takes 3 hours to paint, so \( 3y \) hours total for \( y \) platters.
- A bread basket takes 14 hours to paint, so \( 14z \) hours total for \( z \) bread baskets.
- The workshop has 70 hours available for painting.
\[
2x + 3y + 14z = 70
\]

### Firing time:
- A serving bowl takes 9 hours to fire, so \( 9x \) hours total for \( x \) serving bowls.
- A platter takes 4 hours to fire, so \( 4y \) hours total for \( y \) platters.
- A bread basket takes 7 hours to fire, so \( 7z \) hours total for \( z \) bread baskets.
- The workshop has 109 hours available for firing.
\[
9x + 4y + 7z = 109
\]

We now have the following system of equations:
\[
\begin{aligned}
1. & \quad 3x + 14y + 4z = 92 \\
2. & \quad 2x + 3y + 14z = 70 \\
3. & \quad 9x + 4y + 7z = 109
\]

I'll solve this system of equations to find the values of \( x \), \( y \), and \( z \).The workshop can make:
- 8 serving bowls (\( x = 8 \)),
- 4 platters (\( y = 4 \)),
- 3 bread baskets (\( z = 3 \)).

Would you like more details on how these values were derived or any further clarifications?

Here are five related questions to expand on this problem:
1. How would the solution change if the workshop had more hours available for preparation?
2. What if the preparation time for each bread basket increased by 2 hours?
3. Can the workshop make more items if the firing time per platter was reduced to 2 hours?
4. How many total items can the workshop produce given these time constraints?
5. What is the maximum number of platters the workshop can make if no bowls or bread baskets are made?

**Tip:** In systems of linear equations, altering one constraint often affects the feasibility or quantity of solutions, so recalculating consistently is key to maintaining accuracy.

A ceramics workshop makes serving bowls, platters, and bread baskets to sell at its Winter Festival. A serving bowl takes 3 hours to prepare, 2 hours to paint, and 9 hours to fire. A platter takes 14 hours to prepare, 3 hours to paint, and 4 hours to fire. A bread basket takes 4 hours to prepare, 14 hours to paint, and 7 hours to fire. If the workshop has 92 hours for prep time, 70 hours for painting, and 109 hours for firing, how many of each can be made?

Learn how to solve a system of linear equations to determine the number of serving bowls, platters, and bread baskets that can be produced by a ceramics workshop. This problem involves preparation, painting, and firing constraints and is suitable for students in Grades 10-12.

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