To solve the equation \( 3n - 1 = 2k + 1 \) for \( n \) in terms of \( k \), follow these steps:

1. **Isolate \( n \)** on one side:
   \[
   3n - 1 = 2k + 1
   \]
   Add \( 1 \) to both sides:
   \[
   3n = 2k + 2
   \]

2. **Solve for \( n \)** by dividing both sides by \( 3 \):
   \[
   n = \frac{2k + 2}{3}
   \]

Thus, the solution is:
\[
n = \frac{2k + 2}{3}
\]

### Do you want details or have any questions?
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### Relative Questions:
1. What values of \( k \) make \( n \) an integer in \( n = \frac{2k + 2}{3} \)?
2. How would the solution change if the equation was \( 3n - 1 = 2k - 1 \)?
3. What if \( n \) was isolated in a different way, like subtracting \( 2k \) from both sides?
4. Can you solve \( 3n - 1 = 2k + 1 \) for \( k \) in terms of \( n \)?
5. How does this linear equation graphically compare to the line \( y = mx + b \)?

### Tip:
Always check if your final expression can be simplified or if specific integer solutions exist by checking the divisibility.

Learn how to solve the linear equation 3n - 1 = 2k + 1 to find the value of n in terms of k. This problem involves isolating n, dividing by 3, and ensuring integer solutions for specific values of k. Suitable for high school students and those studying linear equations.

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