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

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.