To solve the rational inequality

$\frac{15x + 2}{5x - 1} \leq 4$

let's work step by step.

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Step 1: Eliminate the denominator

Rearrange the inequality to set up a single rational expression. Start by subtracting 4 from both sides:

$\frac{15x + 2}{5x - 1} - 4 \leq 0$

Rewrite with a common denominator:

$\frac{15x + 2 - 4(5x - 1)}{5x - 1} \leq 0$

Simplify the numerator:

$15x + 2 - 20x + 4 = -5x + 6$

So the inequality becomes:

$\frac{-5x + 6}{5x - 1} \leq 0$

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Step 2: Find critical points

The critical points occur where the numerator and denominator are zero:

1. Numerator: $-5x + 6 = 0$
   Solve for $x$:
   $x = \frac{6}{5}$

2. Denominator: $5x - 1 = 0$
   Solve for $x$:
   $x = \frac{1}{5}$

These points split the number line into intervals for testing.

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Step 3: Analyze intervals

The critical points divide the number line into the following intervals:

1. $(-\infty, \frac{1}{5})$
2. $(\frac{1}{5}, \frac{6}{5})$
3. $(\frac{6}{5}, \infty)$

We test the sign of the rational expression $\frac{-5x + 6}{5x - 1}$ in each interval:

1. Interval $(-\infty, \frac{1}{5})$:
   Pick $x = 0$:
   $\frac{-5(0) + 6}{5(0) - 1} = \frac{6}{-1} = -6 \quad (\text{negative})$

2. Interval $(\frac{1}{5}, \frac{6}{5})$:
   Pick $x = 1$:
   $\frac{-5(1) + 6}{5(1) - 1} = \frac{-5 + 6}{5 - 1} = \frac{1}{4} \quad (\text{positive})$

3. Interval $(\frac{6}{5}, \infty)$:
   Pick $x = 2$:
   $\frac{-5(2) + 6}{5(2) - 1} = \frac{-10 + 6}{10 - 1} = \frac{-4}{9} \quad (\text{negative})$

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Step 4: Determine solution regions

We want the rational expression $\frac{-5x + 6}{5x - 1} \leq 0$. This means the expression should be either negative or zero.

1. Negative in $(-\infty, \frac{1}{5})$ and $(\frac{6}{5}, \infty)$.
2. Zero at $x = \frac{6}{5}$ (where the numerator is zero).
3. Undefined at $x = \frac{1}{5}$ (excluded from the solution).

Thus, the solution set is:

$(-\infty, \frac{1}{5}) \cup [\frac{6}{5}, \infty)$

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Step 5: Graph the solution set

• Draw a real number line.
• Open circle at $x = \frac{1}{5}$ (excluded).
• Closed circle at $x = \frac{6}{5}$ (included).
• Shade regions $(-\infty, \frac{1}{5})$ and $[\frac{6}{5}, \infty)$.

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Final Answer:

The solution set is:

$(-\infty, \frac{1}{5}) \cup [\frac{6}{5}, \infty)$