To solve the equation \( \frac{1}{3}x + 1 = \frac{1}{6} \), we first isolate \( x \).
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Start by subtracting 1 from both sides: \[ \frac{1}{3}x = \frac{1}{6} - 1 \]
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Convert 1 to a fraction with a denominator of 6: \[ 1 = \frac{6}{6} \] Therefore, \[ \frac{1}{6} - \frac{6}{6} = \frac{1 - 6}{6} = \frac{-5}{6} \] Now we have: \[ \frac{1}{3}x = \frac{-5}{6} \]
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To solve for \( x \), multiply both sides by the reciprocal of \( \frac{1}{3} \), which is 3: \[ x = 3 \cdot \frac{-5}{6} \] This simplifies to: \[ x = \frac{-15}{6} = \frac{-5}{2} \]
Now, we want to find the value of \( 2x + 3 \):
- Substitute \( x = \frac{-5}{2} \) into \( 2x + 3 \): \[ 2x + 3 = 2 \left(\frac{-5}{2}\right) + 3 \] Simplifying this gives: \[ 2x + 3 = -5 + 3 = -2 \]
The final answer is: \[ \boxed{-2} \]