1. Solve the inequality $$\frac{1}{x+2} \leq \frac{2}{3x+1}$$ algebraically. 2. First, bring all terms to one side to form a single inequality: $$\frac{1}{x+2} - \frac{2}{3x+1} \leq 0$$ 3. Find a common denominator and combine the fractions: $$\frac{(3x+1) - 2(x+2)}{(x+2)(3x+1)} \leq 0$$ 4. Simplify the numerator: $$3x + 1 - 2x - 4 = x - 3$$ So the inequality becomes: $$\frac{x - 3}{(x+2)(3x+1)} \leq 0$$ 5. Identify critical points where numerator or denominator is zero: - Numerator zero at $x=3$ - Denominator zero at $x=-2$ and $x=-\frac{1}{3}$ 6. These points divide the number line into intervals: $$(-\infty, -2), (-2, -\frac{1}{3}), (-\frac{1}{3}, 3), (3, \infty)$$ 7. Test each interval to determine the sign of the expression: - For $x < -2$, pick $x=-3$: $$\frac{-3 - 3}{(-3+2)(3(-3)+1)} = \frac{-6}{(-1)(-8)} = \frac{-6}{8} < 0$$ - For $-2 < x < -\frac{1}{3}$, pick $x=-1$: $$\frac{-1 - 3}{(-1+2)(3(-1)+1)} = \frac{-4}{(1)(-2)} = \frac{-4}{-2} = 2 > 0$$ - For $-\frac{1}{3} < x < 3$, pick $x=0$: $$\frac{0 - 3}{(0+2)(3(0)+1)} = \frac{-3}{(2)(1)} = -\frac{3}{2} < 0$$ - For $x > 3$, pick $x=4$: $$\frac{4 - 3}{(4+2)(3(4)+1)} = \frac{1}{(6)(13)} > 0$$ 8. The inequality requires the expression to be less than or equal to zero, so include intervals where expression is negative or zero. 9. Check if critical points are included: - At $x=3$, numerator zero, expression equals zero, include $x=3$. - At $x=-2$ or $x=-\frac{1}{3}$, denominator zero, expression undefined, exclude these points. 10. Final solution: $$(-\infty, -2) \cup \left(-\frac{1}{3}, 3\right]$$