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The problem asks to substitute \(\frac{1}{8}\) for \(B\) in Equation (1) and then solve for \(A\).

The equation given is:
\[
0 = A + 2 \left(\frac{1}{8}\right)
\]

### Step-by-Step Solution

1. Substitute \(\frac{1}{8}\) for \(B\):
   \[
   0 = A + 2 \cdot \frac{1}{8}
   \]

2. Simplify the term \(2 \cdot \frac{1}{8}\):
   \[
   2 \cdot \frac{1}{8} = \frac{2}{8} = \frac{1}{4}
   \]

3. Substitute back into the equation:
   \[
   0 = A + \frac{1}{4}
   \]

4. Isolate \(A\) by subtracting \(\frac{1}{4}\) from both sides:
   \[
   A = -\frac{1}{4}
   \]

### Final Answer
\[
A = -\frac{1}{4}
\]

Would you like more details, or do you have any questions?

### Related Questions

1. How would the solution change if we substituted a different value for \(B\)?
2. How can we verify our solution for \(A\) is correct?
3. What are the properties of fractions when dealing with addition and subtraction in equations?
4. Can this equation be solved if \(B\) were an unknown variable rather than a fraction?
5. How would the solution differ if we multiplied both sides of the equation by a common denominator?

### Tip
When dealing with equations involving fractions, consider simplifying each fraction separately before substituting them into the equation for easier calculations.

Substitute \( \frac{1}{8} \) for \( B \) in Equation (1) and solve for \( A \).

Learn how to solve for \( A \) by substituting \( B \) with \( \frac{1}{8} \) in a linear equation. This problem covers basic algebra concepts, focusing on substitution and solving linear equations. Suitable for students in Grades 6-8.

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