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!