The problem asks to find the value of $x$ such that the sum of 6 and one-third of $x$ is one more than twice $x$. We can express this as an equation.

1. Problem Description

The problem asks to find the value of $x$ such that the sum of 6 and one-third of $x$ is one more than twice $x$. We can express this as an equation.

2. Solution Steps

The problem statement can be translated into the following equation:

$6 + \frac{1}{3}x = 2x + 1$

To solve for $x$, we first subtract $\frac{1}{3}x$ from both sides of the equation:

$6 = 2x - \frac{1}{3}x + 1$

$6 = \frac{6}{3}x - \frac{1}{3}x + 1$

$6 = \frac{5}{3}x + 1$

Next, subtract 1 from both sides:

$6 - 1 = \frac{5}{3}x$

$5 = \frac{5}{3}x$

Now, multiply both sides by $\frac{3}{5}$ to isolate $x$:

$5 \cdot \frac{3}{5} = x$

$3 = x$

Thus, $x = 3$.

3. Final Answer

The value of x is

3. So the answer is C. $x=3$.