To determine which equation has a different solution from the rest, we can solve each equation for \( x \).
A.
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\( -17x - 34 = 14 \)
- Add 34 to both sides: \[ -17x = 14 + 34 = 48 \]
- Divide by -17: \[ x = \frac{-48}{17} \]
-
\( -\frac{1}{7}x - \frac{3}{4} = \frac{1}{4} \)
- Add \( \frac{3}{4} \) to both sides: \[ -\frac{1}{7}x = \frac{1}{4} + \frac{3}{4} = 1 \]
- Multiply by -7: \[ x = -7 \]
B.
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\( 12x - 3 = 12 \)
- Add 3 to both sides: \[ 12x = 12 + 3 = 15 \]
- Divide by 12: \[ x = \frac{15}{12} = \frac{5}{4} \]
-
\( \frac{1}{2}x - 3 = \frac{1}{2} \)
- Add 3 to both sides: \[ \frac{1}{2}x = \frac{1}{2} + 3 = \frac{1}{2} + \frac{6}{2} = \frac{7}{2} \]
- Multiply by 2: \[ x = 7 \]
C.
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\( 34x + 5 = -14 \)
- Subtract 5 from both sides: \[ 34x = -14 - 5 = -19 \]
- Divide by 34: \[ x = \frac{-19}{34} \]
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\( \frac{3}{4}x + 5 = -\frac{1}{4} \)
- Subtract 5 from both sides: \[ \frac{3}{4}x = -\frac{1}{4} - 5 = -\frac{1}{4} - \frac{20}{4} = -\frac{21}{4} \]
- Multiply by \(\frac{4}{3}\): \[ x = -\frac{21}{3} = -7 \]
D.
- \( -0.35x - 0.52 = 1.93 \)
- Add 0.52 to both sides: \[ -0.35x = 1.93 + 0.52 = 2.45 \]
- Divide by -0.35: \[ x = \frac{-2.45}{0.35} \approx -7 \]
Summary of Solutions:
- For A: \( x = -7 \) (both parts yield different results)
- For B: \( x = \frac{5}{4} \) and \( x = 7 \) (different results)
- For C: \( x = \frac{-19}{34} \) (from the first part) and \( x = -7 \) (from the second part) (different results)
- For D: \( x \approx -7 \)
After evaluating all equations, the distinct one with consistently different solutions is B, since one of its parts yields \( x = \frac{5}{4} \), which is distinct from the others that lead to -7 or a rational negative value. Thus, the correct answer is:
B.