To determine which of the given equations has a solution different from the rest, we will solve each equation for \( x \).
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Equation 1: \( 34x + 5 = -14 \) \[ 34x = -14 - 5 \ 34x = -19 \ x = -\frac{19}{34} \]
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Equation 2: \( \frac{3}{4}x + 5 = -\frac{1}{4} \) \[ \frac{3}{4}x = -\frac{1}{4} - 5 \ \frac{3}{4}x = -\frac{1}{4} - \frac{20}{4} \ \frac{3}{4}x = -\frac{21}{4} \ x = -\frac{21}{4} \times \frac{4}{3} = -7 \]
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Equation 3: \( -0.35x - 0.52 = 1.93 \) \[ -0.35x = 1.93 + 0.52 = 2.45 \ x = -\frac{2.45}{0.35} \approx -7 \]
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Equation 4: \( -\frac{1}{7}x - \frac{3}{4} = \frac{1}{4} \) \[ -\frac{1}{7}x = \frac{1}{4} + \frac{3}{4} = 1 \ x = -1 \times -7 = 7 \]
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Equation 5: \( 12x - 3 = 12 \) \[ 12x = 12 + 3 \ 12x = 15 \ x = \frac{15}{12} = \frac{5}{4} \]
Now, let's summarize the solutions:
- Equation 1: \( x = -\frac{19}{34} \)
- Equation 2: \( x = -7 \)
- Equation 3: \( x \approx -7 \)
- Equation 4: \( x = 7 \)
- Equation 5: \( x = \frac{5}{4} \)
The solutions for Equations 2 and 3 are approximately equal (\(-7\)), but Equation 4 has a solution of \( 7 \), and Equation 5 has a solution of \( \frac{5}{4} \). However, the solution from Equation 4 (\( 7 \)) is particularly different from the others.
Thus, the equation that has a solution different from the rest is: −17x−34=14 or − \frac{1}{7} x − \frac{3}{4} = \frac{1}{4} leading to \( x = 7 \).
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