To solve the equation \( \frac{1}{3}x + 1 = \frac{1}{6}(2x + 3) \), we will start by eliminating the fraction on the right side.
First, we can distribute \( \frac{1}{6} \) on the right side:
\[ \frac{1}{3}x + 1 = \frac{1}{6} \cdot 2x + \frac{1}{6} \cdot 3 \]
This simplifies to:
\[ \frac{1}{3}x + 1 = \frac{1}{3}x + \frac{1}{2} \]
Next, we can eliminate \( \frac{1}{3}x \) from both sides of the equation:
\[ 1 = \frac{1}{2} \]
This implies that the terms cancel out and we end up with:
\[ 1 \neq \frac{1}{2} \]
Since the equation \( 1 = \frac{1}{2} \) is false, it leads us to conclude that there is no solution for \( x \).
Thus, the final answer is that there is no value of \( x \) that satisfies the given equation.