To determine which of the equations has a solution of \(-\frac{1}{2}\), we can substitute \(-\frac{1}{2}\) into each option and check if the equation holds true.
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Equation: \(x + \frac{1}{2} = -1\)
Substitute \(x = -\frac{1}{2}\):
\[ -\frac{1}{2} + \frac{1}{2} = 0 \quad \text{(not equal to -1)} \]
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Equation: \(x + \frac{1}{2} = -1\)
This is the same as the first equation and has already been checked.
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Equation: \(x - \frac{1}{2} = 1\)
Substitute \(x = -\frac{1}{2}\):
\[ -\frac{1}{2} - \frac{1}{2} = -1 \quad \text{(not equal to 1)} \]
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Equation: \(x + 1 = \frac{1}{2}\)
Substitute \(x = -\frac{1}{2}\):
\[ -\frac{1}{2} + 1 = \frac{1}{2} \quad (This is true) \]
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Equation: \(x - 1 = -\frac{1}{2}\)
Substitute \(x = -\frac{1}{2}\):
\[ -\frac{1}{2} - 1 = -\frac{3}{2} \quad \text{(not equal to -1/2)} \]
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Equation: \(x - 1 = -\frac{1}{2}\)
This is similar to the previous equation and has already been checked.
The only equation that holds true when substituting \(-\frac{1}{2}\) is \(x + 1 = \frac{1}{2}\).
Thus, the equation with a solution of \(-\frac{1}{2}\) is:
\[ \boxed{x + 1 = \frac{1}{2}} \]