Questions: Solve the rational inequality and graph the solution set on a real number line. Express the solution set in interval notation. (3x+1)/(5x-2) ≤ 1

Solution Steps

First, rewrite the inequality \(\frac{3x+1}{5x-2} \leq 1\) by subtracting 1 from both sides:

\[ \frac{3x+1}{5x-2} - 1 \leq 0 \]

Combine the terms into a single fraction:

\[ \frac{3x+1 - (5x-2)}{5x-2} \leq 0 \]

Simplify the numerator:

\[ \frac{3x+1 - 5x + 2}{5x-2} \leq 0 \]

\[ \frac{-2x + 3}{5x-2} \leq 0 \]

Find the critical points by setting the numerator and denominator to zero:

1. Numerator: \(-2x + 3 = 0\)

   \[ x = \frac{3}{2} \]

2. Denominator: \(5x - 2 = 0\)

   \[ x = \frac{2}{5} \]

Test the intervals determined by the critical points \(\left(-\infty, \frac{2}{5}\right)\), \(\left(\frac{2}{5}, \frac{3}{2}\right)\), and \(\left(\frac{3}{2}, \infty\right)\).

1. Choose \(x = 0\) in \(\left(-\infty, \frac{2}{5}\right)\):

   \[ \frac{-2(0) + 3}{5(0) - 2} = \frac{3}{-2} < 0 \]

2. Choose \(x = 1\) in \(\left(\frac{2}{5}, \frac{3}{2}\right)\):

   \[ \frac{-2(1) + 3}{5(1) - 2} = \frac{1}{3} > 0 \]

3. Choose \(x = 2\) in \(\left(\frac{3}{2}, \infty\right)\):

   \[ \frac{-2(2) + 3}{5(2) - 2} = \frac{-1}{8} < 0 \]

The solution set is where the inequality is less than or equal to zero:

• \(\left(-\infty, \frac{2}{5}\right)\) and \(\left(\frac{3}{2}, \infty\right)\)

Since \(\frac{3}{2}\) makes the numerator zero, it is included in the solution set. The point \(\frac{2}{5}\) is not included because it makes the denominator zero.

Final Answer