### Instruction: Find all real numbers $x$ such that \[3 \le \frac{x}{2x-5} < 8.\](Give your answer in interval notation.).### Response: To solve this inequality, I need to find the values of $x$ that make the fraction between 3 and 8, inclusive.
I can start by splitting the inequality into two parts: \[3 \le \frac{x}{2x-5} \quad \text{and} \quad \frac{x}{2x-5} < 8.\]
For each part, I can try to isolate $x$ by cross-multiplying and simplifying.
For the first part, I get \[3(2x-5) \le x \implies 6x-15 \le x \implies 5x \le 15 \implies x \le 3.\]
For the second part, I get \[\frac{x}{2x-5} < 8 \implies x < 8(2x-5) \implies x < 16x-40 \implies -15x < -40 \implies x > \frac{8}{3}.\]
So, I have two conditions for $x$: \[x \le 3 \quad \text{and} \quad x > \frac{8}{3}.\]
However, I need to be careful, because there is a potential problem with the fraction: it is undefined when the denominator is zero.
So, I also need to avoid the value of $x$ that makes $2x-5 = 0$, which is $x = \frac{5}{2}$.
Therefore, the solution set for $x$ is the intersection of the intervals \[\left(\frac{8}{3}, \frac{5}{2}\right) \cup \left(\frac{5}{2}, 3\right].\]
In interval notation, this is \[\left(\frac{8}{3}, \frac{5}{2}\right) \cup \left(\frac{5}{2}, 3\right].\]