Solve the following equation for BB.
2, equals, minus, 3, f, plus, one fifth, R, B
2=
−3f+ 
5
1
​
 RB

The equation you provided seems to be:

\[
2 = -3f + \frac{1}{5} RB
\]

We are solving for \(B\). Let's go through the steps to isolate \(B\).

### Step 1: Move the term involving \(B\) to one side
Subtract \(-3f\) from both sides:

\[
2 + 3f = \frac{1}{5} RB
\]

### Step 2: Eliminate the fraction by multiplying both sides by 5
Multiply everything by 5 to get rid of the denominator:

\[
5(2 + 3f) = RB
\]

Simplifying:

\[
10 + 15f = RB
\]

### Step 3: Solve for \(B\)
Now, divide both sides by \(R\) to isolate \(B\):

\[
B = \frac{10 + 15f}{R}
\]

### Final Answer:
\[
B = \frac{10 + 15f}{R}
\]

Let me know if you'd like more details or if you have any questions!

Here are 5 related questions for further exploration:
1. How do we solve equations when \(B\) appears in more than one term?
2. What if \(R = 0\)? How does that change the solution?
3. Can you factor out common terms before solving for \(B\)?
4. How does changing the constants (like 2 or 3) affect the solution process?
5. How can you apply this type of equation in physics or economics?

**Tip:** Always check if a variable can be factored out across terms before solving—it simplifies the process.

Solve the following equation for B: 2 = -3f + (1/5)RB

This solution walks you through how to solve for B in the equation 2 = -3f + (1/5)RB. It uses algebraic concepts, including isolating variables and solving linear equations, making it ideal for students in Grades 9-12.

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