To solve for \( k \) in the equation \( 20 - 6 + 4k = 2 - 2k \), we need to simplify and isolate the variable \( k \). Here’s a step-by-step solution:

1. **Simplify both sides of the equation:**

\[
20 - 6 + 4k = 2 - 2k
\]

Simplify the constants on the left side:

\[
14 + 4k = 2 - 2k
\]

2. **Combine like terms by moving all terms involving \( k \) to one side and constants to the other:**

To move \( -2k \) to the left side, add \( 2k \) to both sides:

\[
14 + 4k + 2k = 2
\]

Simplify:

\[
14 + 6k = 2
\]

Now, move the constant \( 14 \) to the right side by subtracting \( 14 \) from both sides:

\[
6k = 2 - 14
\]

Simplify:

\[
6k = -12
\]

3. **Solve for \( k \):**

To isolate \( k \), divide both sides by 6:

\[
k = \frac{-12}{6}
\]

Simplify:

\[
k = -2
\]

So, the solution for \( k \) is \( \mathbf{-2} \).

Would you like more details or have any questions?

Here are some related questions:

1. What if the equation was \(20 - 6 + 4k = 5 - 2k\)? What would \(k\) be then?
2. How would you solve the equation if there was a different constant term on the right side?
3. Can you verify the solution \(k = -2\) by substituting it back into the original equation?
4. What steps would you take if \(k\) was a fractional number in the equation?
5. How would the equation change if there were additional terms with \(k^2\) or higher powers of \(k\)?

**Tip:** Always perform the same operation on both sides of the equation to maintain equality.

Learn how to solve the equation 20 - 6 + 4k = 2 - 2k to find the value of k. This problem involves linear equations and step-by-step solution methods. Suitable for students in Grades 7-9.

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