To determine which equation has a solution different from the rest, we need to solve both equations for \(x\).
- For the first equation: \[ \frac{1}{2}x - 3 = \frac{1}{2} \]
Add 3 to both sides: \[ \frac{1}{2}x = \frac{1}{2} + 3 \] \[ \frac{1}{2}x = \frac{1}{2} + \frac{6}{2} = \frac{7}{2} \]
Now, multiply both sides by 2 to solve for \(x\): \[ x = 7 \]
- For the second equation: \[ \frac{3}{4}x + 5 = -\frac{1}{4} \]
Subtract 5 from both sides: \[ \frac{3}{4}x = -\frac{1}{4} - 5 \] \[ \frac{3}{4}x = -\frac{1}{4} - \frac{20}{4} = -\frac{21}{4} \]
Now, multiply both sides by \(\frac{4}{3}\) to solve for \(x\): \[ x = -\frac{21}{4} \times \frac{4}{3} = -\frac{21}{3} = -7 \]
Now we summarize the solutions:
- First equation: \(x = 7\)
- Second equation: \(x = -7\)
Since the solutions are \(7\) and \(-7\), the equation that has a solution different from the other is the second equation: \[ \frac{3}{4}x + 5 = -\frac{1}{4} \]