Let's solve the given rational inequality:

$\frac{4x + 5}{3x - 1} > 4$

Step-by-step Solution:

1. Move all terms to one side: To simplify the inequality, let's subtract 4 from both sides:

   $\frac{4x + 5}{3x - 1} - 4 > 0$

2. Express the left side with a common denominator: The expression can be written with a common denominator:

   $\frac{4x + 5}{3x - 1} - \frac{4(3x - 1)}{3x - 1} > 0$

   Simplifying the second fraction:

   $\frac{4x + 5 - 12x + 4}{3x - 1} > 0$

   This simplifies to:

   $\frac{-8x + 9}{3x - 1} > 0$

3. Find critical points: The critical points occur where the numerator or denominator is zero. Let's find these:

   • Numerator: $-8x + 9 = 0$
     Solving for $x$:
     $x = \frac{9}{8}$

   • Denominator: $3x - 1 = 0$
     Solving for $x$:
     $x = \frac{1}{3}$

4. Determine sign changes: We now test the sign of the expression $\frac{-8x + 9}{3x - 1}$ in the intervals determined by the critical points $x = \frac{1}{3}$ and $x = \frac{9}{8}$.

   The intervals are:

   • $(-\infty, \frac{1}{3})$
   • $(\frac{1}{3}, \frac{9}{8})$
   • $(\frac{9}{8}, \infty)$

   We can test points in each interval to determine where the expression is positive.

   • For $x = 0$ (in $(-\infty, \frac{1}{3})$): $\frac{-8(0) + 9}{3(0) - 1} = \frac{9}{-1} = -9 \quad \text{(negative)}$

   • For $x = \frac{1}{2}$ (in $(\frac{1}{3}, \frac{9}{8})$): $\frac{-8\left(\frac{1}{2}\right) + 9}{3\left(\frac{1}{2}\right) - 1} = \frac{-4 + 9}{1.5 - 1} = \frac{5}{0.5} = 10 \quad \text{(positive)}$

   • For $x = 2$ (in $(\frac{9}{8}, \infty)$): $\frac{-8(2) + 9}{3(2) - 1} = \frac{-16 + 9}{6 - 1} = \frac{-7}{5} \quad \text{(negative)}$

5. Solution in interval notation: The solution is where the expression is positive, which occurs in the interval $\left( \frac{1}{3}, \frac{9}{8} \right)$.

Thus, the solution is:

$\left( \frac{1}{3}, \frac{9}{8} \right)$

Would you like a detailed explanation of any specific step?

Here are 5 related questions to explore further:

1. How do we find common denominators in rational expressions?
2. Why do we test values in different intervals to determine signs?
3. What does the critical point of a rational inequality signify?
4. Can this method be applied to solve other types of inequalities?
5. What is the importance of excluding points where the denominator is zero?

Tip: Always check for points that make the denominator zero since they are not part of the solution set!