Question

Identify the pair of linear equations that have the same solution set.

4 = k/2 + 5 and 4j - 14/3 + 5 = 3
4k - 5 = 2k and 8j - 2 = 3j
3 (m + 1) = 10 and 6n + 3 = 6 - n
2 (3g + 5) - 2g = 2 (4 - g) and -36h/6 = 2

Answer 1

To identify the pair of linear equations that have the same solution set, we'll solve each pair of equations and see if they yield the same solution.

1. **First Pair:**
- \( 4 = \frac{k}{2} + 5 \)
- Rearranging gives: \(\frac{k}{2} = 4 - 5 \rightarrow \frac{k}{2} = -1 \rightarrow k = -2\)

- \( 4j - \frac{14}{3} + 5 = 3 \)
- Rearranging gives: \( 4j - \frac{14}{3} + \frac{15}{3} = 3 \rightarrow 4j + \frac{1}{3} = 3 \rightarrow 4j = 3 - \frac{1}{3}\)
- \( 4j = \frac{9}{3} - \frac{1}{3} = \frac{8}{3} \rightarrow j = \frac{2}{3}\)

2. **Second Pair:**
- \( 4k - 5 = 2k \)
- Rearranging gives: \( 4k - 2k = 5 \rightarrow 2k = 5 \rightarrow k = \frac{5}{2}\)

- \( 8j - 2 = 3j \)
- Rearranging gives: \( 8j - 3j = 2 \rightarrow 5j = 2 \rightarrow j = \frac{2}{5}\)

3. **Third Pair:**
- \( 3(m + 1) = 10 \)
- Expanding gives: \( 3m + 3 = 10 \rightarrow 3m = 10 - 3 \rightarrow 3m = 7 \rightarrow m = \frac{7}{3}\)

- \( 6n + 3 = 6 - n \)
- Rearranging gives: \( 6n + n = 6 - 3 \rightarrow 7n = 3 \rightarrow n = \frac{3}{7}\)

4. **Fourth Pair:**
- \( 2(3g + 5) - 2g = 2(4 - g) \)
- Expanding gives: \( 6g + 10 - 2g = 8 - 2g \rightarrow 4g + 10 = 8 - 2g \rightarrow 4g + 2g = 8 - 10 \rightarrow 6g = -2 \rightarrow g = -\frac{1}{3}\)

- \( -\frac{36h}{6} = 2 \)
- Simplifying gives: \( -6h = 2 \rightarrow h = -\frac{1}{3}\)

Now we can summarize the solutions:

1. First Pair: \( k = -2, j = \frac{2}{3} \)
2. Second Pair: \( k = \frac{5}{2}, j = \frac{2}{5} \)
3. Third Pair: \( m = \frac{7}{3}, n = \frac{3}{7} \)
4. Fourth Pair: \( g = -\frac{1}{3}, h = -\frac{1}{3} \)

The only pair that has the same solution is the **Fourth Pair**:
- \( g = -\frac{1}{3} \) and \( h = -\frac{1}{3} \)

**So the answer is the fourth pair: \(2(3g + 5) - 2g = 2(4 - g)\) and \(-\frac{36h}{6} = 2\)**.