Let's solve the equation \( \frac{1}{4}b + 1 = -\frac{1}{2}b + 3 \) for \( b \).

### Step 1: Eliminate the fractions by finding a common denominator.
To eliminate the fractions, multiply every term by 4 (the least common denominator between 4 and 2).

\[
4 \times \left(\frac{1}{4}b + 1\right) = 4 \times \left(-\frac{1}{2}b + 3\right)
\]

This simplifies to:

\[
b + 4 = -2b + 12
\]

### Step 2: Move the terms involving \( b \) to one side.
Add \( 2b \) to both sides to bring the \( b \) terms together:

\[
b + 2b + 4 = 12
\]

This simplifies to:

\[
3b + 4 = 12
\]

### Step 3: Solve for \( b \).
Subtract 4 from both sides:

\[
3b = 8
\]

Finally, divide by 3:

\[
b = \frac{8}{3}
\]

So, the solution is \( b = \frac{8}{3} \) or approximately 2.67.

Would you like more details or have any questions? 

Here are five related questions to further explore the topic:
1. How do you solve equations with fractions step by step?
2. Can you solve \( \frac{3}{5}b + 2 = \frac{1}{5}b + 4 \)?
3. What is the advantage of eliminating fractions early in an equation?
4. How do you handle equations where fractions have different denominators?
5. How do you verify if the solution \( b = \frac{8}{3} \) is correct?

**Tip:** When dealing with fractions in equations, eliminating them early by multiplying through by the least common denominator can simplify the process significantly.

Learn how to solve the equation 1/4b + 1 = -1/2b + 3 step-by-step. This problem involves equations with fractions, suitable for Grades 7-9. Understand the process of eliminating fractions and solving linear equations.

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