In a, pottery factory, a machine takes to 3 minutes make a glass and 2 minutes to make a plate. The material for a glass cost $0.25 and the material for a plate cost $0.20. If the machine runs for 10 hours and exactly $50 is used up for material. How many glass and plates can be produced. 

Let's denote:

- $ x $ = number of glasses produced
- $ y $ = number of plates produced

We know the following:

1. **Time constraint**: The machine runs for 10 hours, which is equivalent to $ 10 \times 60 = 600 $ minutes. The time taken to make each glass is 3 minutes, and the time taken to make each plate is 2 minutes. Therefore, the total time used is:

\[
3x + 2y = 600
\]

2. **Cost constraint**: The cost of material for each glass is $0.25, and the cost of material for each plate is $0.20. The total cost of materials used is $50. Thus, the cost equation is:

\[
0.25x + 0.20y = 50
\]

Now we have a system of two equations:

\[
\begin{aligned}
1. & \quad 3x + 2y = 600 \\
2. & \quad 0.25x + 0.20y = 50
\end{aligned}
\]

### Solving the system of equations:

1. First, multiply the second equation by 100 to eliminate the decimals:

\[
25x + 20y = 5000
\]

So the system becomes:

\[
\begin{aligned}
3x + 2y &= 600 \\
25x + 20y &= 5000
\end{aligned}
\]

2. To eliminate $ y $, multiply the first equation by 10:

\[
30x + 20y = 6000
\]

Now subtract the second equation from this:

\[
(30x + 20y) - (25x + 20y) = 6000 - 5000
\]

This simplifies to:

\[
5x = 1000
\]

Solving for $ x $:

\[
x = 200
\]

3. Substitute $ x = 200 $ into the first equation:

\[
3(200) + 2y = 600
\]

\[
600 + 2y = 600
\]

\[
2y = 0
\]

\[
y = 0
\]

### Conclusion:

The machine can produce **200 glasses** and **0 plates** in 10 hours, using exactly $50 of material.

---

Would you like more details or clarifications on this? Here are 5 related questions you might explore:

1. How would the solution change if the material cost of a plate increased to $0.30?
2. What if the machine only ran for 8 hours instead of 10?
3. How does the time constraint affect the number of plates that can be produced?
4. What if the production times for glasses and plates were swapped?
5. How would changing both the cost and production time of plates affect the result?

**Tip**: Always check for special cases when solving systems of equations, such as solutions where one variable is zero.

In a pottery factory, a machine takes 3 minutes to make a glass and 2 minutes to make a plate. The material for a glass costs $0.25 and the material for a plate costs $0.20. If the machine runs for 10 hours and exactly $50 is used up for material, how many glasses and plates can be produced?

Learn how to solve a system of equations for a pottery factory problem where a machine produces glasses and plates under time and material cost constraints. The solution involves solving linear equations and understanding algebraic methods. Suitable for Grades 8-10.

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