To determine which pair of linear equations has the same solution set, we can solve each pair for their respective variables and compare if they yield the same results.
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First Pair:
- \(4k - 5 = 2k\)
- Rearranging gives \(4k - 2k = 5\) → \(2k = 5\) → \(k = \frac{5}{2}\)
- \(8j - 2 = 3j\)
- Rearranging gives \(8j - 3j = 2\) → \(5j = 2\) → \(j = \frac{2}{5}\)
Different solutions.
- \(4k - 5 = 2k\)
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Second Pair:
- \(4 = k^2 + 5\)
- Rearranging gives \(k^2 = 4 - 5\) → \(k^2 = -1\) → No real solutions.
- \(4j - \frac{14}{3} + 5 = 3\)
- Converting constants gives \(4j - \frac{14}{3} + \frac{15}{3} = 3\) → \(4j + \frac{1}{3} = 3\) → \(4j = 3 - \frac{1}{3}\) → \(4j = \frac{9 - 1}{3} = \frac{8}{3}\) → \(j = \frac{2}{3}\)
Different solutions.
- \(4 = k^2 + 5\)
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Third Pair:
- \(3(m + 1) = 10\)
- Expanding gives \(3m + 3 = 10\) → \(3m = 7\) → \(m = \frac{7}{3}\)
- \(6n + 3 = 6 - n\)
- Rearranging gives \(6n + n = 6 - 3\) → \(7n = 3\) → \(n = \frac{3}{7}\)
Different solutions.
- \(3(m + 1) = 10\)
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Fourth Pair:
- \(2(3g + 5) - 2g = 2(4 - g)\)
- Expanding gives \(6g + 10 - 2g = 8 - 2g\) → \(4g + 10 = 8 - 2g\) → \(6g + 10 = 8\) → \(6g = -2\) → \(g = -\frac{1}{3}\)
- \(-\frac{36h}{6} = 2\) simplifies to \(-6h = 2\) → \(h = -\frac{1}{3}\)
Same solution, both result in \(-\frac{1}{3}\).
- \(2(3g + 5) - 2g = 2(4 - g)\)
The solution that yields the same result for its variables is the fourth pair:
2(3g + 5) - 2g = 2(4 - g) and -36h/6 = 2.