We are tasked with solving the inequality:

\[
\frac{x+4}{2x+1} \geq 3x
\]

### Step 1: Move all terms to one side
We start by moving all terms to one side of the inequality:

\[
\frac{x+4}{2x+1} - 3x \geq 0
\]

### Step 2: Combine the terms
To combine these terms, we need a common denominator. The common denominator between \(\frac{x+4}{2x+1}\) and \(3x\) is \(2x+1\). Rewrite \(3x\) with the denominator \(2x+1\):

\[
\frac{x+4}{2x+1} - \frac{3x(2x+1)}{2x+1} \geq 0
\]

Now, combine the fractions:

\[
\frac{(x+4) - 3x(2x+1)}{2x+1} \geq 0
\]

### Step 3: Simplify the numerator
Expand the terms in the numerator:

\[
(x + 4) - 3x(2x+1) = x + 4 - (6x^2 + 3x)
\]
\[
= x + 4 - 6x^2 - 3x
\]
\[
= -6x^2 - 2x + 4
\]

So, the inequality becomes:

\[
\frac{-6x^2 - 2x + 4}{2x+1} \geq 0
\]

### Step 4: Solve the inequality
Now, we need to find where the expression \(\frac{-6x^2 - 2x + 4}{2x+1} \geq 0\) holds.

1. **Find the roots of the numerator**: Solve \(-6x^2 - 2x + 4 = 0\).

   Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = -6\), \(b = -2\), and \(c = 4\):

   \[
   x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(-6)(4)}}{2(-6)}
   \]
   \[
   x = \frac{2 \pm \sqrt{4 + 96}}{-12} = \frac{2 \pm \sqrt{100}}{-12} = \frac{2 \pm 10}{-12}
   \]
   \[
   x = \frac{2 + 10}{-12} = \frac{12}{-12} = -1, \quad x = \frac{2 - 10}{-12} = \frac{-8}{-12} = \frac{2}{3}
   \]

   So, the roots of the numerator are \(x = -1\) and \(x = \frac{2}{3}\).

2. **Find the root of the denominator**: Solve \(2x + 1 = 0\).

   \[
   2x = -1 \quad \Rightarrow \quad x = -\frac{1}{2}
   \]

### Step 5: Analyze the sign of the expression
The critical points are \(x = -1\), \(x = -\frac{1}{2}\), and \(x = \frac{2}{3}\). These points divide the real line into intervals. We will test each interval to determine where the expression is non-negative.

- For \(x < -1\)
- For \(-1 < x < -\frac{1}{2}\)
- For \(-\frac{1}{2} < x < \frac{2}{3}\)
- For \(x > \frac{2}{3}\)

Evaluating the signs in each interval:

- \((-∞, -1)\): Negative.
- \((-1, -\frac{1}{2})\): Positive.
- \((- \frac{1}{2}, \frac{2}{3})\): Negative (denominator changes sign).
- \((\frac{2}{3}, ∞)\): Positive.

### Step 6: Conclusion
The solution to the inequality is:

\[
x \in [-1, -\frac{1}{2}) \cup \left( \frac{2}{3}, \infty \right)
\]

Would you like to go into more detail or clarify any part? 

Here are five related questions to extend this topic:
1. How would you approach solving a strict inequality \(\frac{x+4}{2x+1} > 3x\)?
2. What if the numerator were a cubic equation instead of a quadratic one?
3. How can you use graphing methods to solve this inequality?
4. What happens if the denominator has a higher degree than the numerator?
5. How can you apply this method to inequalities involving absolute values?

**Tip**: Always check the denominator in rational inequalities to avoid dividing by zero!

Learn how to solve the inequality (x+4)/(2x+1) ≥ 3x with a step-by-step solution. This problem involves simplifying rational expressions and solving quadratic inequalities, using key algebraic concepts and the quadratic formula. Suitable for Grades 10-12.

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