To determine which pairs of linear equations have the same solution set, we need to manipulate each equation to see if we can arrive at the same final format for each corresponding pair.
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Equation Set 1:
- \( 2(3g + 5) - 2g = 2(4 - g) \)
Expanding both sides: \[ 6g + 10 - 2g = 8 - 2g \] \[ 4g + 10 = 8 - 2g \] Adding \( 2g \) to both sides: \[ 6g + 10 = 8 \] Subtracting \( 10 \) from both sides: \[ 6g = -2 \] \[ g = -\frac{1}{3} \]
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Equation Set 2:
- \( 4k - 5 = 2k \)
- \( 8j - 2 = 3j \)
For \( 4k - 5 = 2k \): \[ 4k - 2k = 5 \] \[ 2k = 5 \] \[ k = \frac{5}{2} \]
For \( 8j - 2 = 3j \): \[ 8j - 3j = 2 \] \[ 5j = 2 \] \[ j = \frac{2}{5} \]
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Equation Set 3:
- \( 3(m + 1) = 10 \)
- \( 6n + 3 = 6 - n \)
For \( 3(m + 1) = 10 \): \[ 3m + 3 = 10 \] \[ 3m = 7 \] \[ m = \frac{7}{3} \]
For \( 6n + 3 = 6 - n \): \[ 6n + n = 3 \] \[ 7n = 3 \] \[ n = \frac{3}{7} \]
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Equation Set 4:
- \( 4 = k^2 + 5 \)
- \( 4j - 143 + 5 = 3 \)
For \( 4 = k^2 + 5 \): \[ k^2 = -1 \] (No real solution)
For \( 4j - 143 + 5 = 3 \): \[ 4j - 138 = 3 \] \[ 4j = 141 \] \[ j = \frac{141}{4} \]
After calculating all pairs, we can see:
- Pair 1 has \( g = -\frac{1}{3} \).
- Pair 2 has diverse results (\( k = \frac{5}{2} \), \( j = \frac{2}{5} \)).
- Pair 3 also differs (\( m = \frac{7}{3} \), \( n = \frac{3}{7} \)).
- Pair 4 has no solution for \( k \) but has a defined solution for \( j \).
Conclusion: The pairs of equations do not have the same solution set. However, they represent different linear equations and yield distinct solutions.
In the cases you provided, none of the presented pairs have been shown to have the same solution set based on the evaluations and factorizations above. If you have a specific question about which ones may resemble each other closely or require further analysis, let me know!