To solve for \( a \), we can use the equation \( FM = 8a + 1 \) and the information given: \( FG = 42 \). However, it seems we are missing how \( FM \) and \( FG \) are related. Assuming \( FM = FG \), you can substitute 42 for \( FM \) in the equation.

1. Start with the equation:
   \[
   FM = 8a + 1
   \]
2. Substitute \( FM = 42 \):
   \[
   42 = 8a + 1
   \]
3. Solve for \( a \):
   \[
   42 - 1 = 8a
   \]
   \[
   41 = 8a
   \]
   \[
   a = \frac{41}{8}
   \]
   \[
   a = 5.125
   \]

Thus, \( a = 5.125 \).

Would you like more details or have any questions?

Here are 5 related questions:
1. What happens if \( FM \) and \( FG \) represent different values?
2. Can you solve for \( a \) if \( FM = 0 \)?
3. How would you solve if the equation was quadratic instead?
4. What are the units of \( FM \) and \( FG \) in this context?
5. How does changing the constant in the equation impact \( a \)?

**Tip:** Always verify how variables relate to each other before solving for unknowns.

Solve the linear equation FM = 8a + 1 with the given value FG = 42. Follow a step-by-step approach to find the value of a, using basic algebraic techniques. Suitable for students in Grades 7-8.

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