To determine which pair of linear equations has the same solution set, we need to simplify each pair and see if they can be transformed into equivalent equations.
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First pair: \[ 4k - 5 = 2k \] Simplifying gives: \[ 4k - 2k = 5 \quad \Rightarrow \quad 2k = 5 \quad \Rightarrow \quad k = \frac{5}{2} \]
\[ 8j - 2 = 3j \] Simplifying gives: \[ 8j - 3j = 2 \quad \Rightarrow \quad 5j = 2 \quad \Rightarrow \quad j = \frac{2}{5} \]
These are different solutions.
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Second pair: \[ 4 = k^2 + 5 \] Rearranging gives: \[ k^2 = 4 - 5 \quad \Rightarrow \quad k^2 = -1 \] No real solution exists for \(k\).
\[ 4j - 14 + 5 = 3 \] Simplifying gives: \[ 4j - 9 = 3 \quad \Rightarrow \quad 4j = 12 \quad \Rightarrow \quad j = 3 \]
Different solutions again (no real solution for \(k\)).
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Third pair: \[ 3(m+1) = 10 \quad \Rightarrow \quad 3m + 3 = 10 \quad \Rightarrow \quad 3m = 7 \quad \Rightarrow \quad m = \frac{7}{3} \]
\[ 6n + 3 = 6 - n \] Rearranging gives: \[ 6n + n = 6 - 3 \quad \Rightarrow \quad 7n = 3 \quad \Rightarrow \quad n = \frac{3}{7} \]
Different solutions again.
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Fourth pair: \[ 2(3g + 5) - 2g = 2(4 - g) \] Expanding and simplifying: \[ 6g + 10 - 2g = 8 - 2g \quad \Rightarrow \quad 4g + 10 = 8 - 2g \quad \Rightarrow \quad 6g = -2 \quad \Rightarrow \quad g = -\frac{1}{3} \]
\[ -36h = 6 \quad \Rightarrow \quad h = -\frac{1}{6} \]
Different solutions again.
In conclusion, after examining all options, none of the pairs of equations provided have the same solution set. If you have further equations or pairs, please provide them for evaluation!