To determine which two equations have the same solution, we can solve each equation and compare their solutions.
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Equation 1: \( 6z - 10 = 50 \)
- Add 10 to both sides: \( 6z = 60 \)
- Divide by 6: \( z = 10 \)
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Equation 2: \( 2(5z + 4) = 8x + 28 \)
- Divide everything by 2: \( 5z + 4 = 4x + 14 \)
- Subtract 4 from both sides: \( 5z = 4x + 10 \)
- Solve for z: \( z = \frac{4x + 10}{5} \)
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Equation 3: \( \frac{x}{2} + 4 = 5x - 10 \)
- Subtract \( \frac{x}{2} \) from both sides: \( 4 + 10 = 5x - \frac{x}{2} \)
- Combine like terms: \( 14 = \frac{10x - x}{2} \) \( 28 = 9x \)
- Divide by 9: \( x = \frac{28}{9} \)
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Equation 4: \( \frac{5x}{9} = 3 \)
- Multiply both sides by 9: \( 5x = 27 \)
- Divide by 5: \( x = \frac{27}{5} \)
Now, let's summarize the solutions:
- \( z = 10 \) for Equation 1
- \( z = \frac{4x + 10}{5} \) for Equation 2 (depends on \( x \))
- \( x = \frac{28}{9} \) for Equation 3
- \( x = \frac{27}{5} \) for Equation 4
From the analysis, it appears that:
- Equation 1 (\( 6z - 10 = 50 \)) has a solution of \( z = 10 \).
Equation 3 and Equation 4 have different \( x \) solutions and don't provide the same \( z \) solution.
Thus, only Equation 2 can potentially match with Equation 1 depending on solving it for specific relations.
However, if we weren't looking at the forms too closely, we should note that the key on matching directly remains:
Final Pairing: To answer the question, we pair accordingly:
- Responses:
- \( 6z - 10 = 50 \) has same solution as \( 2 (5z + 4) = 8x + 28 \)
Drag as needed:
Response input area has the same solution as \( 6z - 10 = 50 \)
Response input area \( 2 (5z + 4) = 8x + 28 \).