To solve the equation \((8 + y)h = C\) for a specific variable, we can rearrange the equation depending on which variable you need to isolate.

1. **Solving for \( h \):**
   \[
   h = \frac{C}{8 + y}
   \]

2. **Solving for \( y \):**
   \[
   y = \frac{C}{h} - 8
   \]

3. **Solving for \( C \):**
   \[
   C = (8 + y)h
   \]

These are the different ways to rearrange the equation depending on which variable you're solving for. Let me know if you want more details!

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Here are 5 related questions:

1. What happens if \( h = 0 \) in the equation \((8 + y)h = C\)?
2. How does the value of \( y \) change if \( h \) increases while \( C \) remains constant?
3. If \( C = 0 \), what values of \( h \) and \( y \) satisfy the equation?
4. Can you find a real-world example where an equation like this might be used?
5. How would you graph the relationship between \( h \) and \( y \) if \( C \) is constant?

**Tip:** Always check for undefined expressions when dividing, especially if the denominator can potentially be zero.

Discover how to solve the linear equation (8 + y)h = C by isolating different variables like h, y, and C. This problem helps students understand basic algebraic concepts, including variable isolation, and is suitable for students in Grades 6-8.

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