To determine which equation has a different solution from the rest, we can solve each equation one by one.
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Equation 1: \( \frac{3}{4}x + 5 = -\frac{1}{4} \)
\(\frac{3}{4}x = -\frac{1}{4} - 5\)
\(\frac{3}{4}x = -\frac{1}{4} - \frac{20}{4} = -\frac{21}{4}\)
\(x = -\frac{21}{4} \cdot \frac{4}{3} = -\frac{21}{3} = -7\)
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Equation 2: \(-0.35x - 0.52 = 1.93\)
\(-0.35x = 1.93 + 0.52\)
\(-0.35x = 2.45\)
\(x = \frac{2.45}{-0.35} \approx -7\)
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Equation 3: \(\frac{1}{2}x - 3 = \frac{1}{2}\)
\(\frac{1}{2}x = \frac{1}{2} + 3\)
\(\frac{1}{2}x = \frac{1}{2} + \frac{6}{2} = \frac{7}{2}\)
\(x = 7\)
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Equation 4: \(-\frac{1}{7}x - \frac{3}{4} = \frac{1}{4}\)
\(-\frac{1}{7}x = \frac{1}{4} + \frac{3}{4}\)
\(-\frac{1}{7}x = 1\)
\(x = -7\)
Now we have the solutions for each equation:
- Equation 1: \(x = -7\)
- Equation 2: \(x \approx -7\)
- Equation 3: \(x = 7\)
- Equation 4: \(x = -7\)
The equation that has a solution different from the rest is Equation 3, which has the solution \(x = 7\). The others yield \(x \approx -7\).