Identify the pair of linear equations that have the same solution set.(1 point)

To find the pair of linear equations that have the same solution set, we can simplify each pair and check if they lead to equivalent equations.

1. **First pair:**
- \( 2(3g + 5) - 2g = 2(4 - g) \)
- Expanding both sides:
- Left: \( 6g + 10 - 2g = 4g + 10 \)
- Right: \( 8 - 2g \)
- So, \( 4g + 10 = 8 - 2g \)
- Rearranging gives \( 4g + 2g = 8 - 10\)
- Which simplifies to \( 6g = -2 \), so \( g = -\frac{1}{3} \)

- \( -\frac{36h}{6} = 2 \) simplifies to:
- \( -6h = 2 \)
- So, \( h = -\frac{1}{3} \)

- Not equivalent as the variables are different.

2. **Second pair:**
- \( 3(m + 1) = 10 \)
- Expanding gives \( 3m + 3 = 10 \)
- Simplifying gives \( 3m = 7 \) or \( m = \frac{7}{3} \)

- \( 6n + 3 = 6 - n \)
- Rearranging gives \( 6n + n = 6 - 3 \)
- Or \( 7n = 3 \), so \( n = \frac{3}{7} \)

- Not equivalent.

3. **Third pair:**
- \( 4k - 5 = 2k \)
- Simplifying gives \( 2k = 5 \), or \( k = \frac{5}{2} \)

- \( 8j - 2 = 3j \)
- Rearranging gives \( 8j - 3j = 2 \)
- Or \( 5j = 2 \), so \( j = \frac{2}{5} \)

- Not equivalent.

4. **Fourth pair:**
- \( 4 = \frac{k}{2} + 5 \)
- Rearranging gives \( \frac{k}{2} = 4 - 5 \)
- Or \( \frac{k}{2} = -1\), so \( k = -2 \)

- \( \frac{4j - 14}{3} + 5 = 3 \)
- Rearranging gives \( \frac{4j - 14}{3} = 3 - 5 \)
- Or \( \frac{4j - 14}{3} = -2 \), leading to \( 4j - 14 = -6 \)
- Simplifying gives \( 4j = 8 \), so \( j = 2 \)

- Not equivalent.

Conclusion:
None of the pairs given have the same solution set based on the computed results. Consequently, none of the options are correct as pairs of equations that have matching solution sets.