1. **State the problem:** Solve the inequality $$\frac{4x + 1}{5x - 3} \leq 2$$. 2. **Rewrite the inequality:** Subtract 2 from both sides to get a single rational expression: $$\frac{4x + 1}{5x - 3} - 2 \leq 0$$ 3. **Find a common denominator and combine:** $$\frac{4x + 1 - 2(5x - 3)}{5x - 3} \leq 0$$ 4. **Simplify the numerator:** $$4x + 1 - 10x + 6 = -6x + 7$$ So the inequality becomes: $$\frac{-6x + 7}{5x - 3} \leq 0$$ 5. **Determine critical points:** Set numerator and denominator equal to zero: - Numerator zero: $$-6x + 7 = 0 \Rightarrow x = \frac{7}{6}$$ - Denominator zero: $$5x - 3 = 0 \Rightarrow x = \frac{3}{5}$$ (excluded from domain) 6. **Test intervals around critical points:** The number line is divided into three intervals: - $$(-\infty, \frac{3}{5})$$ - $$\left(\frac{3}{5}, \frac{7}{6}\right)$$ - $$\left(\frac{7}{6}, \infty\right)$$ Choose test points in each interval to check the sign of the expression: - For $$x=0$$ (in $$(-\infty, \frac{3}{5})$$): numerator $$-6(0)+7=7>0$$, denominator $$5(0)-3=-3<0$$, fraction is negative (positive/negative = negative) $\leq 0$ true. - For $$x=1$$ (in $$\left(\frac{3}{5}, \frac{7}{6}\right)$$): numerator $$-6(1)+7=1>0$$, denominator $$5(1)-3=2>0$$, fraction positive $\leq 0$ false. - For $$x=2$$ (in $$\left(\frac{7}{6}, \infty\right)$$): numerator $$-6(2)+7=-5<0$$, denominator $$5(2)-3=7>0$$, fraction negative $\leq 0$ true. 7. **Include points where fraction equals zero:** Numerator zero at $$x=\frac{7}{6}$$, so include this point. 8. **Exclude points where denominator is zero:** At $$x=\frac{3}{5}$$, denominator zero, so exclude this point. 9. **Final solution:** $$x \in \left(-\infty, \frac{3}{5}\right) \cup \left[\frac{7}{6}, \infty\right)$$ **Answer:** $$\boxed{x \in (-\infty, \frac{3}{5}) \cup [\frac{7}{6}, \infty)}$$