To identify the pair of linear equations that have the same solution set, we can solve each pair to see if they yield equivalent equations.
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For the first pair:
- \(3(m+1)=10\) \[ 3m + 3 = 10 \implies 3m = 7 \implies m = \frac{7}{3} \]
- \(6n + 3 = 6 - n\) \[ 6n + n = 6 - 3 \implies 7n = 3 \implies n = \frac{3}{7} \]
These do not have the same solution.
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For the second pair:
- \(2(3g+5)-2g=2(4-g)\) \[ 6g + 10 - 2g = 8 - 2g \] \[ 6g - 2g + 2g = 8 - 10 \implies 6g = -2 \implies g = -\frac{1}{3} \]
- \(-\frac{36h}{6}=2\) \[ -6h = 2 \implies h = -\frac{1}{3} \]
These have the same solution.
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For the third pair:
- \(4 = \frac{k}{2} + 5\) \[ 4 - 5 = \frac{k}{2} \implies -1 = \frac{k}{2} \implies k = -2 \]
- \(4j - \frac{14}{3} + 5 = 3\) \[ 4j + 5 - 3 = \frac{14}{3} \implies 4j + 2 = \frac{14}{3} \] \[ 4j = \frac{14}{3} - 2 = \frac{14}{3} - \frac{6}{3} = \frac{8}{3} \implies j = \frac{2}{3} \]
These do not have the same solution.
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For the fourth pair:
- \(4k - 5 = 2k\) \[ 4k - 2k = 5 \implies 2k = 5 \implies k = \frac{5}{2} \]
- \(8j - 2 = 3j\) \[ 8j - 3j = 2 \implies 5j = 2 \implies j = \frac{2}{5} \]
These do not have the same solution.
Thus, the pair of linear equations that have the same solution set is:
2(3g+5)−2g=2(4−g) and −36h/6=2.
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