To determine which pair of linear equations has the same solution set, we need to solve each pair of equations and check if they yield equivalent solutions.
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Equations: \[ 3(m + 1) = 10 \] Simplifying: \[ 3m + 3 = 10 \implies 3m = 7 \implies m = \frac{7}{3} \]
Second equation: \[ 6n + 3 = 6 - n \] Rearranging: \[ 6n + n = 6 - 3 \implies 7n = 3 \implies n = \frac{3}{7} \] (These equations are not equivalent in solutions.)
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Equations: \[ 4 = \frac{k}{2} + 5 \] Simplifying: \[ \frac{k}{2} = 4 - 5 \implies \frac{k}{2} = -1 \implies k = -2 \]
Second equation: \[ \frac{4j - 14}{3} + 5 = 3 \] Rearranging: \[ \frac{4j - 14}{3} = 3 - 5 \implies \frac{4j - 14}{3} = -2 \] Multiplying through by 3: \[ 4j - 14 = -6 \implies 4j = 8 \implies j = 2 \] (These equations are not equivalent in solutions.)
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Equations: \[ 2(3g + 5) - 2g = 2(4 - g) \] Expanding: \[ 6g + 10 - 2g = 8 - 2g \implies 4g + 10 = 8 - 2g \] Rearranging: \[ 4g + 2g = 8 - 10 \implies 6g = -2 \implies g = -\frac{1}{3} \]
Second equation: \[ \frac{-36h}{6} = 2 \] Simplifying: \[ -6h = 2 \implies h = -\frac{1}{3} \] (These equations are equivalent in solutions.)
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Equations: \[ 4k - 5 = 2k \] Rearranging: \[ 4k - 2k = 5 \implies 2k = 5 \implies k = \frac{5}{2} \]
Second equation: \[ 8j - 2 = 3j \] Rearranging: \[ 8j - 3j = 2 \implies 5j = 2 \implies j = \frac{2}{5} \] (These equations are not equivalent in solutions.)
Conclusion: The only pair of equations that have the same solution set is: \[ 2(3g + 5) - 2g = 2(4 - g) \quad \text{and} \quad \frac{-36h}{6} = 2 \] So, the final answer is: \[ \boxed{(2(3g+5)-2g=2(4-g) \text{ and } \frac {-36h}{6}=2)} \]
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