Let's solve the equation:

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

### Step 1: Distribute the 3 on the right-hand side
Multiply 3 by both terms inside the parentheses:

\[
\frac{1}{4}k = 3 \cdot \left(-\frac{1}{4}k\right) + 3 \cdot 3
\]

This simplifies to:

\[
\frac{1}{4}k = -\frac{3}{4}k + 9
\]

### Step 2: Add \(\frac{3}{4}k\) to both sides to get all the \(k\)-terms on one side:

\[
\frac{1}{4}k + \frac{3}{4}k = 9
\]

### Step 3: Combine like terms on the left-hand side:

\[
\frac{4}{4}k = 9
\]

Simplifying:

\[
k = 9
\]

### Final Answer:
The solution to the equation is \(k = 9\).

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

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Here are 5 related questions you might find useful:

1. How would you solve the equation if the right-hand side had a different coefficient?
2. What happens if the constants on both sides differ by a fraction?
3. Can you verify the solution by substituting \(k = 9\) back into the original equation?
4. How does solving this equation change if fractions are removed early on by multiplying both sides by a common denominator?
5. How do we handle equations with variables on both sides involving more complex fractions?

**Tip:** Always double-check your solution by plugging the value of the variable back into the original equation to verify it!

Learn how to solve the linear equation \( \frac{1}{4}k = 3(-\frac{1}{4}k + 3) \) with a detailed step-by-step solution. This problem covers key algebraic concepts, including the distributive property and properties of equality. Suitable for students in Grades 8-10.

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