To determine which pair of linear equations has the same solution set, we need to simplify each pair of equations and see if they lead to identical results.
1. **First Pair:**
- **Equation 1:** \(4k - 5 = 2k\)
\[
4k - 2k = 5 \implies 2k = 5 \implies k = \frac{5}{2}
\]
- **Equation 2:** \(8j - 2 = 3j\)
\[
8j - 3j = 2 \implies 5j = 2 \implies j = \frac{2}{5}
\]
The solutions \(k = \frac{5}{2}\) and \(j = \frac{2}{5}\) are different.
2. **Second Pair:**
- **Equation 1:** \(4 = \frac{k}{2} + 5\)
\[
\frac{k}{2} = 4 - 5 \implies \frac{k}{2} = -1 \implies k = -2
\]
- **Equation 2:** \(\frac{4j - 14}{3} + 5 = 3\)
\[
\frac{4j - 14}{3} = 3 - 5 \implies \frac{4j - 14}{3} = -2 \implies 4j - 14 = -6 \implies 4j = 8 \implies j = 2
\]
The solutions \(k = -2\) and \(j = 2\) are different.
3. **Third Pair:**
- **Equation 1:** \(2(3g + 5) - 2g = 2(4 - g)\)
\[
6g + 10 - 2g = 8 - 2g \implies 4g + 10 = 8 \implies 4g = -2 \implies g = -\frac{1}{2}
\]
- **Equation 2:** \(-\frac{36h}{6} = 2\)
\[
-6h = 2 \implies h = -\frac{1}{3}
\]
The solutions \(g = -\frac{1}{2}\) and \(h = -\frac{1}{3}\) are different.
4. **Fourth Pair:**
- **Equation 1:** \(3(m + 1) = 10\)
\[
3m + 3 = 10 \implies 3m = 7 \implies m = \frac{7}{3}
\]
- **Equation 2:** \(6n + 3 = 6 - n\)
\[
7n + 3 = 6 \implies 7n = 3 \implies n = \frac{3}{7}
\]
The solutions \(m = \frac{7}{3}\) and \(n = \frac{3}{7}\) are different.
After evaluating all pairs, none of the pairs have the same solution set. Therefore, none of the groups provided result in the same solution for both equations.