Find all real numbers $x$ such that \[3 \le \frac{x}{2x-5} < 8.\](Give your answer in interval notation.)
We work on the two parts of the given inequality separately. First, $3 \le \frac{x}{2x-5}$ is equivalent to \[0 \le \frac{x}{2x-5} - 3 = \frac{x - 3(2x-5)}{2x-5} = \frac{-5x + 15}{2x-5}.\]Making a sign table, we have:  \begin{tabular}{c|cc|c} &$-5x+15$ &$2x-5$ &$\frac{-5x+15}{2x-5}$ \\ \hline$x<\frac{5}{2}$ &$+$&$-$&$-$\\ [.1cm]$\frac{5}{2}<x<3$ &$+$&$+$&$+$\\ [.1cm]$x>3$ &$-$&$+$&$-$\\ [.1cm]\end{tabular}Therefore, the inequality holds when $\tfrac52 < x < 3,$ as well as the endpoint $x = 3,$ which makes the right-hand side zero. The solution set to the first inequality is $(\tfrac52, 3].$

Second, $\frac{x}{2x-5} < 8$ is equivalent to \[\frac{x}{2x-5} - 8 = \frac{x - 8(2x-5)}{2x-5} = \frac{-15x + 40}{2x-5} < 0.\]Making another sign table, we have: \begin{tabular}{c|cc|c} &$-15x+40$ &$2x-5$ &$\frac{-15x+40}{2x-5}$ \\ \hline$x<\frac{5}{2}$ &$+$&$-$&$-$\\ [.1cm]$\frac{5}{2}<x<\frac{8}{3}$ &$+$&$+$&$+$\\ [.1cm]$x>\frac{8}{3}$ &$-$&$+$&$-$\\ [.1cm]\end{tabular}It follows that the inequality holds when either $x < \tfrac52$ or $x > \tfrac83.$

The intersection of this solution set with $(\tfrac52, 3]$ is $\boxed{(\tfrac83, 3]},$ which is the solution set for both inequalities combined.
(\tfrac{8}{3},3]
Level 4
Intermediate Algebra
Problem: Find all real numbers $x$ such that \[3 \le \frac{x}{2x-5} < 8.\](Give your answer in interval notation.)
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the inequality \(3 \le \frac{x}{2x-5} < 8\), we need to break it into two parts: \(3 \le \frac{x}{2x-5}\) and \(\frac{x}{2x-5} < 8\).

Starting with the first part, \(3 \le \frac{x}{2x-5}\):
1. Multiply both sides by \(2x - 5\), noting that we must consider the sign of \(2x - 5\). This gives us two cases: \(2x - 5 > 0\) and \(2x - 5 < 0\).
2. For \(2x - 5 > 0\) (i.e., \(x > \frac{5}{2}\)), we have:
   \[
   3(2x - 5) \le x \implies 6x - 15 \le x \implies 5x \le 15 \implies x \le 3.
   \]
   Thus, for this case, \(x\) must satisfy \(\frac{5}{2} < x \le 3\).

3. For \(2x - 5 < 0\) (i.e., \(x < \frac{5}{2}\)), we have:
   \[
   3(2x - 5) \ge x \implies 6x - 15 \ge x \implies 5x \ge 15 \implies x \ge 3.
   \]
   This case yields no valid solutions since \(x\) cannot be both less than \(\frac{5}{2}\) and greater than or equal to \(3\).

Next, we analyze the second part, \(\frac{x}{2x-5} < 8\):
1. Again, multiply both sides by \(2x - 5\), considering the sign of \(2x - 5\).
2. For \(2x - 5 > 0\) (i.e., \(x > \frac{5}{2}\)), we have:
   \[
   x < 8(2x - 5) \implies x < 16x - 40 \implies 15x < 40 \implies x < \frac{8}{3}.
   \]
   Thus, for this case, \(x\) must satisfy \(\frac{5}{2} < x < \frac{8}{3}\).

3. For \(2x - 5 < 0\) (i.e., \(x < \frac{5}{2}\)), we have:
   \[
   x > 8(2x - 5) \implies x > 16x - 40 \implies -15x > -40 \implies x < \frac{8}{3}.
   \]
   This case also yields no valid solutions since \(x\) cannot be both less than \(\frac{5}{2}\) and less than \(\frac{8}{3}\).

Combining the results from both parts, we find that \(x\) must satisfy:
\[
\frac{5}{2} < x < \frac{8}{3} \quad \text{and} \quad x \le 3.
\]
The intersection of these intervals gives us the final solution. The valid interval is \((\frac{8}{3}, 3]\).<|END_OF_LATENT|> The final answer is $(\tfrac{8}{3},3]$.