We are tasked with solving the rational inequality:

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

Step 1: Move all terms to one side of the inequality.

First, we subtract $\frac{-5}{x}$ from both sides to get:

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

This simplifies to:

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

Step 2: Find a common denominator.

The least common denominator (LCD) between $3x - 1$ and $x$ is $x(3x - 1)$. Now, rewrite both fractions with the LCD:

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

Now simplify the numerator:

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

Expand the numerator:

$\frac{4x + 15x - 5}{x(3x - 1)} > 0$

Simplify:

$\frac{19x - 5}{x(3x - 1)} > 0$

Step 3: Find the critical points.

The critical points occur when the numerator or denominator is zero.

1. Numerator: $19x - 5 = 0$

   Solving for $x$:

   $19x = 5 \quad \Rightarrow \quad x = \frac{5}{19}$

2. Denominator: $x(3x - 1) = 0$

   Set each factor to zero:

   $x = 0 \quad \text{or} \quad 3x - 1 = 0$

   Solving for $x$ in the second equation:

   $3x = 1 \quad \Rightarrow \quad x = \frac{1}{3}$

So, the critical points are $x = 0$, $x = \frac{1}{3}$, and $x = \frac{5}{19}$.

Step 4: Analyze the sign of the expression in each interval.

The critical points divide the real number line into four intervals:

• $(-\infty, 0)$
• $(0, \frac{5}{19})$
• $\left(\frac{5}{19}, \frac{1}{3}\right)$
• $\left(\frac{1}{3}, \infty\right)$

Now, test a value from each interval in the expression $\frac{19x - 5}{x(3x - 1)}$:

1. For $x = -1$ in $(- \infty, 0)$:

   $\frac{19(-1) - 5}{(-1)(3(-1) - 1)} = \frac{-19 - 5}{-1(-3 - 1)} = \frac{-24}{4} = -6$

   This is negative.

2. For $x = \frac{1}{10}$ in $(0, \frac{5}{19})$:

   $\frac{19\left(\frac{1}{10}\right) - 5}{\left(\frac{1}{10}\right)(3\left(\frac{1}{10}\right) - 1)} = \frac{\frac{19}{10} - 5}{\frac{1}{10}(\frac{3}{10} - 1)}$

   This simplifies to a negative value.

3. For $x = \frac{1}{4}$ in $\left(\frac{5}{19}, \frac{1}{3}\right)$:

   Testing will show this is positive.

4. For $x = 1$ in $\left(\frac{1}{3}, \infty\right)$:

   Testing will show this is positive.

Step 5: Write the solution in interval notation.

The expression is positive in the intervals:

$\left(\frac{5}{19}, \frac{1}{3}\right) \cup \left(\frac{1}{3}, \infty\right)$

Thus, the solution set in interval notation is:

$\boxed{\left(\frac{5}{19}, \frac{1}{3}\right) \cup \left(\frac{1}{3}, \infty\right)}$

Would you like more details on any of these steps?

Here are some related questions for further practice:

1. Solve $\frac{3}{x - 2} \leq \frac{-1}{x + 1}$.
2. Find the solution set for $\frac{2x + 3}{x - 1} > 4$.
3. Solve $\frac{x^2}{x + 1} \geq 2$.
4. Determine the solution for $\frac{5}{2x - 3} < \frac{7}{x + 2}$.
5. Solve $\frac{4}{x - 5} \geq \frac{9}{x + 3}$.

Tip: Always be careful when multiplying or dividing inequalities by negative values—this will flip the inequality sign.