Let's solve each equation to find out which one has a different solution.
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Equation 1: \( \frac{3}{4}x + 5 = -\frac{1}{4} \)
Subtract 5 from both sides: \[ \frac{3}{4}x = -\frac{1}{4} - 5 \] Convert 5 to a fraction: \[ 5 = \frac{20}{4} \quad \Rightarrow \quad \frac{3}{4}x = -\frac{1}{4} - \frac{20}{4} = -\frac{21}{4} \] Multiply both sides by \(\frac{4}{3}\): \[ x = -\frac{21}{4} \cdot \frac{4}{3} = -\frac{21}{3} = -7 \]
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Equation 2: \( -\frac{1}{7}x - \frac{3}{4} = \frac{1}{4} \)
Add \(\frac{3}{4}\) to both sides: \[ -\frac{1}{7}x = \frac{1}{4} + \frac{3}{4} \] \[ -\frac{1}{7}x = 1 \] Multiply both sides by -7: \[ x = -7 \]
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Equation 3: \( -0.35x - 0.52 = 1.93 \)
Add \(0.52\) to both sides: \[ -0.35x = 1.93 + 0.52 = 2.45 \] Divide by -0.35: \[ x = \frac{2.45}{-0.35} \approx -7 \]
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Equation 4: \( \frac{1}{2}x - 3 = \frac{1}{2} \)
Add 3 to both sides: \[ \frac{1}{2}x = \frac{1}{2} + 3 \quad \Rightarrow \quad \frac{1}{2}x = \frac{1}{2} + \frac{6}{2} = \frac{7}{2} \] Multiply both sides by 2: \[ x = 7 \]
Now, let's summarize the solutions:
- Equation 1 gives \( x = -7 \)
- Equation 2 gives \( x = -7 \)
- Equation 3 gives \( x \approx -7 \)
- Equation 4 gives \( x = 7 \)
Conclusion: The fourth equation (\( \frac{1}{2}x - 3 = \frac{1}{2} \)) has a solution \( x = 7 \), while the others all yield \( x = -7 \). Thus, the equation with a solution different from the rest is:
4. \( \frac{1}{2}x - 3 = \frac{1}{2} \)
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