Solve the rational inequality and graph the solution set on a real number line. Express the solution set in interval notation. \( \frac{17 x-4}{5 x-2} \leq 4 \)

To solve the rational inequality: \[ \frac{17x - 4}{5x - 2} \leq 4 \] **Step 1: Bring all terms to one side** \[ \frac{17x - 4}{5x - 2} - 4 \leq 0 \] **Step 2: Combine the terms over a common denominator** \[ \frac{17x - 4 - 4(5x - 2)}{5x - 2} \leq 0 \] \[ \frac{17x - 4 - 20x + 8}{5x - 2} \leq 0 \] \[ \frac{-3x + 4}{5x - 2} \leq 0 \] **Step 3: Identify critical points** - **Numerator zero:** \(-3x + 4 = 0 \Rightarrow x = \frac{4}{3}\) - **Denominator zero:** \(5x - 2 = 0 \Rightarrow x = \frac{2}{5}\) (This point is excluded from the solution as it makes the denominator zero.) **Step 4: Determine intervals** The critical points divide the real number line into three intervals: 1. \((-∞, \frac{2}{5})\) 2. \((\frac{2}{5}, \frac{4}{3})\) 3. \((\frac{4}{3}, ∞)\) **Step 5: Test each interval** 1. **For \(x < \frac{2}{5}\) (e.g., \(x = 0\)):** \[ \frac{-3(0) + 4}{5(0) - 2} = \frac{4}{-2} = -2 \leq 0 \quad \text{(True)} \] 2. **For \(\frac{2}{5} < x < \frac{4}{3}\) (e.g., \(x = 1\)):** \[ \frac{-3(1) + 4}{5(1) - 2} = \frac{1}{3} > 0 \quad \text{(False)} \] 3. **For \(x > \frac{4}{3}\) (e.g., \(x = 2\)):** \[ \frac{-3(2) + 4}{5(2) - 2} = \frac{-2}{8} = -\frac{1}{4} \leq 0 \quad \text{(True)} \] **Step 6: Compile the solution** The inequality holds for: - \(x < \frac{2}{5}\) - \(x \geq \frac{4}{3}\) **Final Answer in Interval Notation:** \[ (-\infty,\ \tfrac{2}{5})\ \cup\ [\tfrac{4}{3},\ \infty) \] **Graph:** - A number line with: - An open circle at \(x = \frac{2}{5}\) (excluded). - A closed circle at \(x = \frac{4}{3}\) (included). - Shaded regions to the left of \(\frac{2}{5}\) and to the right of \(\frac{4}{3}\). --- **Answer:** After simplifying, all real numbers below 2⁄5 or at least 4⁄3 satisfy the inequality. In interval notation, ( –∞, 2/5 ) ∪ [ 4/3, ∞ ) A number line graph shows an open circle at 2⁄5 and shading left of it, a closed circle at 4⁄3, and shading to the right of 4⁄3.