Question: What is the range of $y = \frac{2x+1}{3x-6}$?
1. **State the problem:** Find the range of the function $$y = \frac{2x+1}{3x-6}$$.
2. **Recall the domain restriction:** The denominator cannot be zero, so $$3x - 6 \neq 0 \Rightarrow x \neq 2$$.
3. **Find the range:** To find the range, solve for $x$ in terms of $y$:
$$y = \frac{2x+1}{3x-6}$$
Multiply both sides by $3x - 6$:
$$y(3x - 6) = 2x + 1$$
Distribute $y$:
$$3xy - 6y = 2x + 1$$
Group terms with $x$ on one side:
$$3xy - 2x = 6y + 1$$
Factor out $x$:
$$x(3y - 2) = 6y + 1$$
Divide both sides by $3y - 2$ (noting $3y - 2 \neq 0$):
$$x = \frac{6y + 1}{3y - 2}$$
4. **Check for values of $y$ that make denominator zero:**
Set denominator $3y - 2 = 0$:
$$3y - 2 = 0 \Rightarrow y = \frac{2}{3}$$
At $y = \frac{2}{3}$, $x$ is undefined, so $y = \frac{2}{3}$ is not in the range.
5. **Conclusion:** The range is all real numbers except $$\boxed{\frac{2}{3}}$$.
**Answer choice:** D. all real numbers except $\frac{2}{3}$