To determine which pair of linear equations have the same solution set, let's solve each pair of equations one by one.
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First pair:
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\( 4k - 5 = 2k \)
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Rearranging, we get \( 4k - 2k = 5 \) → \( 2k = 5 \) → \( k = \frac{5}{2} \)
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\( 8j - 2 = 3j \)
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Rearranging, we get \( 8j - 3j = 2 \) → \( 5j = 2 \) → \( j = \frac{2}{5} \)
(Different solutions)
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Second pair:
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\( 2(3g + 5) - 2g = 2(4 - g) \)
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Expanding: \( 6g + 10 - 2g = 8 - 2g \)
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Combining terms: \( 4g + 10 = 8 - 2g \) → \( 4g + 2g = 8 - 10 \) → \( 6g = -2 \) → \( g = -\frac{1}{3} \)
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\( -\frac{36h}{6} = 2 \) simplifies to \( -6h = 2 \) → \( h = -\frac{1}{3} \)
(Same solution)
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Third pair:
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\( 4 = \frac{k}{2} + 5 \)
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Rearranging gives: \( \frac{k}{2} = 4 - 5 \) → \( \frac{k}{2} = -1 \) → \( k = -2 \)
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\( \frac{4j - 14}{3} + 5 = 3 \)
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Rearranging gives: \( \frac{4j - 14}{3} = 3 - 5 \) → \( \frac{4j - 14}{3} = -2 \) → \( 4j - 14 = -6 \) → \( 4j = 8 \) → \( j = 2 \)
(Different solutions)
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Fourth pair:
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\( 3(m + 1) = 10 \)
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Expanding gives: \( 3m + 3 = 10 \) → \( 3m = 7 \) → \( m = \frac{7}{3} \)
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\( 6n + 3 = 6 - n \)
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Rearranging gives: \( 6n + n = 6 - 3 \) → \( 7n = 3 \) → \( n = \frac{3}{7} \)
(Different solutions)
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After evaluating all pairs, the only pair with the same solution set is:
2(3g + 5) - 2g = 2(4 - g) and −36h/6 = 2.
3 answers