To identify which pair of linear equations have the same solution set, we can solve each pair and see if they yield the same solution for their respective variables.
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First Pair:
- \( 4 = \frac{k}{2} + 5 \)
- \( 4j - \frac{14}{3} + 5 = 3 \)
For the first equation: \[ 4 - 5 = \frac{k}{2} \ -1 = \frac{k}{2} \ k = -2 \]
For the second equation: \[ 4j - \frac{14}{3} + 5 = 3 \ 4j + \frac{1}{3} = 3 \ 4j = 3 - \frac{1}{3} \ 4j = \frac{9}{3} - \frac{1}{3} = \frac{8}{3} \ j = \frac{2}{3} \]
So, this pair does not have the same solution set.
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Second Pair:
- \( 3(m + 1) = 10 \)
- \( 6n + 3 = 6 - n \)
For the first equation: \[ 3m + 3 = 10 \ 3m = 7 \ m = \frac{7}{3} \]
For the second equation: \[ 6n + 3 = 6 - n \ 6n + n = 6 - 3 \ 7n = 3 \ n = \frac{3}{7} \]
So, this pair does not have the same solution set.
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Third Pair:
- \( 2(3g + 5) - 2g = 2(4 - g) \)
- \( -\frac{36h}{6} = 2 \)
For the first equation: \[ 6g + 10 - 2g = 8 - 2g \ 6g + 2g + 2g = 8 - 10 \ 6g = -2 \ g = -\frac{1}{3} \]
For the second equation: \[ -6h = 2 \ h = -\frac{1}{3} \]
This pair has the same solution set \( g = -\frac{1}{3} \) and \( h = -\frac{1}{3} \).
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Fourth Pair:
- \( 4k - 5 = 2k \)
- \( 8j - 2 = 3j \)
For the first equation: \[ 4k - 2k = 5 \ 2k = 5 \ k = \frac{5}{2} \]
For the second equation: \[ 8j - 3j = 2 \ 5j = 2 \ j = \frac{2}{5} \]
So, this pair does not have the same solution set.
Thus, the pair of linear equations that have the same solution set is: 2(3g + 5) - 2g = 2(4 - g) and -36h/6 = 2.