Questions: Solve the rational inequality and graph the solution set on a real number line. (3x+2)/(5x-1) <= 1 Solve the inequality. What is the solution set? Select the correct choice below and, A. The solution set is (Simplify your answer, Type your answer in interval notation. Type an exact B. The solution set is the empty set.

Solution Steps

To solve the inequality \(\frac{3x + 2}{5x - 1} \leq 1\), we first set up the inequality: \[ \frac{3x + 2}{5x - 1} \leq 1 \]

Subtract 1 from both sides: \[ \frac{3x + 2}{5x - 1} - 1 \leq 0 \]

Combine the terms over a common denominator: \[ \frac{3x + 2 - (5x - 1)}{5x - 1} \leq 0 \]

Simplify the numerator: \[ \frac{3x + 2 - 5x + 1}{5x - 1} \leq 0 \] \[ \frac{-2x + 3}{5x - 1} \leq 0 \]

The critical points are where the numerator and denominator are zero: \[ -2x + 3 = 0 \quad \Rightarrow \quad x = \frac{3}{2} \] \[ 5x - 1 = 0 \quad \Rightarrow \quad x = \frac{1}{5} \]

We test the intervals \((-\infty, \frac{1}{5})\), \((\frac{1}{5}, \frac{3}{2})\), and \((\frac{3}{2}, \infty)\).

1. For \(x < \frac{1}{5}\), choose \(x = 0\): \[ \frac{-2(0) + 3}{5(0) - 1} = \frac{3}{-1} = -3 \quad (\text{negative}) \]

2. For \(\frac{1}{5} < x < \frac{3}{2}\), choose \(x = 1\): \[ \frac{-2(1) + 3}{5(1) - 1} = \frac{1}{4} \quad (\text{positive}) \]

3. For \(x > \frac{3}{2}\), choose \(x = 2\): \[ \frac{-2(2) + 3}{5(2) - 1} = \frac{-1}{9} \quad (\text{negative}) \]

The inequality \(\frac{-2x + 3}{5x - 1} \leq 0\) is satisfied in the intervals where the expression is negative or zero. The critical points \(x = \frac{1}{5}\) and \(x = \frac{3}{2}\) are included because the inequality is \(\leq\).

Thus, the solution set is: \[ (-\infty, \frac{1}{5}] \cup [\frac{3}{2}, \infty) \]

Final Answer