Solve the rational inequality. Write the solution set in interval notation.
StartFraction 4 Over 3 x minus 1 EndFraction greater than StartFraction negative 5 Over x EndFraction


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.

Solve the rational inequality. Write the solution set in interval notation. \(\frac{4}{3x - 1} > \frac{-5}{x}\)

Learn how to solve the rational inequality 4/(3x - 1) > -5/x with a comprehensive step-by-step solution. This problem covers algebraic concepts such as rational inequalities, critical points analysis, and interval notation. Suitable for students in Grades 9-12.

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