To determine which pair of linear equations have the same solution set, we need to simplify and solve each pair of equations.
Option A:
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\(4k - 5 = 2k\)
- Rearranging gives \(4k - 2k = 5\), or \(2k = 5\), leading to \(k = \frac{5}{2}\).
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\(8j - 2 = 3j\)
- Rearranging gives \(8j - 3j = 2\), or \(5j = 2\), leading to \(j = \frac{2}{5}\).
Since \(k = \frac{5}{2}\) and \(j = \frac{2}{5}\) are different, Option A does not have the same solution.
Option B:
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\(3(m + 1) = 10\)
- Expanding gives \(3m + 3 = 10\). Rearranging yields \(3m = 7\), so \(m = \frac{7}{3}\).
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\(6n + 3 = 6 - n\)
- Rearranging gives \(6n + n = 6 - 3\), or \(7n = 3\), leading to \(n = \frac{3}{7}\).
Since \(m \neq n\), Option B does not have the same solution.
Option C:
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\(4 = \frac{k}{2} + 5\)
- To isolate \(k\), rearranging gives \(\frac{k}{2} = 4 - 5\), which simplifies to \(\frac{k}{2} = -1\). Therefore, \(k = -2\).
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\( \frac{4j - 14}{3} + 5 = 3\)
- Multiplying the entire equation by \(3\) gives \(4j - 14 + 15 = 9\), which simplifies to \(4j + 1 = 9\). Rearranging yields \(4j = 8\), so \(j = 2\).
Since \(k \neq j\), Option C does not have the same solution.
Option D:
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\(2(3g + 5) - 2g = 2(4 - g)\)
- Expanding gives \(6g + 10 - 2g = 8 - 2g\). Combining like terms gives \(4g + 10 = 8 - 2g\). Rearranging gives \(6g = -2\), so \(g = -\frac{1}{3}\).
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\(-\frac{36h}{6} = 2\)
- This simplifies to \(-6h = 2\), resulting in \(h = -\frac{1}{3}\).
Since \(g = h = -\frac{1}{3}\), Option D has the same solution.
The correct answer is Option D.