To solve the compound inequality:

\[
3x + \frac{1}{2} \leq 2x + \frac{5}{2} < \frac{x}{4} + 25
\]

We will break it into two parts and solve each inequality separately:

### 1. Solve \(3x + \frac{1}{2} \leq 2x + \frac{5}{2}\)

Subtract \(2x\) from both sides:

\[
x + \frac{1}{2} \leq \frac{5}{2}
\]

Subtract \(\frac{1}{2}\) from both sides:

\[
x \leq \frac{5}{2} - \frac{1}{2} = \frac{4}{2} = 2
\]

So, the solution to the first part is:

\[
x \leq 2
\]

### 2. Solve \(2x + \frac{5}{2} < \frac{x}{4} + 25\)

Subtract \(2x\) from both sides:

\[
\frac{5}{2} < \frac{x}{4} - 2x + 25
\]

Combine the \(x\) terms on the right-hand side:

\[
\frac{5}{2} < \frac{x}{4} - \frac{8x}{4} + 25
\]

\[
\frac{5}{2} < \frac{-7x}{4} + 25
\]

Subtract 25 from both sides:

\[
\frac{5}{2} - 25 < \frac{-7x}{4}
\]

Convert 25 to a fraction with denominator 2:

\[
\frac{5}{2} - \frac{50}{2} < \frac{-7x}{4}
\]

\[
\frac{-45}{2} < \frac{-7x}{4}
\]

Multiply both sides by \(-4/7\) to solve for \(x\) (remember to reverse the inequality sign when multiplying by a negative):

\[
x < \frac{4}{7} \times \frac{45}{2} = \frac{180}{14} = \frac{90}{7} \approx 12.86
\]

### Combined Solution

The solution is the intersection of \(x \leq 2\) and \(x < \frac{90}{7}\). Since \(x \leq 2\) is more restrictive:

\[
\boxed{x \leq 2}
\]

This is the solution to the compound inequality.

Would you like more details or have any questions? 

Here are some related questions to further explore:
1. How do you solve inequalities involving fractions?
2. What happens if you multiply or divide an inequality by a negative number?
3. How do you graph the solution of a compound inequality?
4. Can you provide another example of solving a compound inequality?
5. How do you solve inequalities with absolute values?

**Tip:** When solving compound inequalities, remember to treat each part separately and then find the intersection of the solutions.

Learn how to solve the compound inequality 3x+1/2 ≤ 2x+5/2 < x/4+25 with a detailed step-by-step solution. This problem covers inequalities involving fractions and is suitable for advanced high school students. Gain insights into solving complex inequalities with clear explanations.

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