To determine which of the given equations has infinitely many solutions, we need to check if any of them simplify to an identity (i.e., they are true for all values of the variable).
Let's analyze each option:
A: \((8y - 1) = 2y - \frac{1}{8}(8y - 1)\)
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Start by simplifying the right side: \[\frac{1}{8}(8y - 1) = y - \frac{1}{8}\] So, the equation becomes: \[8y - 1 = 2y - (y - \frac{1}{8})\] \[8y - 1 = 2y - y + \frac{1}{8}\] \[8y - 1 = y + \frac{1}{8}\]
Now, move \(y\) to the left: \[8y - y = 1 - \frac{1}{8}\] \[7y = \frac{8}{8} - \frac{1}{8} = \frac{7}{8}\] \[y = \frac{1}{8}\]
This equation has only one solution (\(y = \frac{1}{8}\)), not infinitely many.
B: \(8x + 24 = -4(2x - 6)\)
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Expand the right side: \[8x + 24 = -8x + 24\]
Now, add \(8x\) to both sides: \[16x + 24 = 24\]
Subtract 24 from both sides: \[16x = 0\] \[x = 0\]
This equation also has only one solution (\(x = 0\)), not infinitely many.
C: \(-3(6x + 4) + 30x = 12(x - 1)\)
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Expand both sides: \[-18x - 12 + 30x = 12x - 12\]
Combine like terms: \[12x - 12 = 12x - 12\]
Since both sides are equal for all \(x\), this equation has infinitely many solutions.
D: \(0.5(4y + 16) = 2y - 8\)
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Start by simplifying: \[2y + 8 = 2y - 8\]
Subtract \(2y\) from both sides: \[8 = -8\]
This is a contradiction, indicating that there are no solutions.
Thus, the equation with infinitely many solutions is C: \(-3(6x + 4) + 30x = 12(x - 1)\).