1. Solve $$\frac{2x}{x+6} \geq 1$$ Step 1: Move all terms to one side: $$\frac{2x}{x+6} - 1 \geq 0$$ Step 2: Write with common denominator: $$\frac{2x - (x+6)}{x+6} \geq 0\implies \frac{2x - x - 6}{x+6} \geq 0$$ Simplify numerator: $$\frac{x - 6}{x+6} \geq 0$$ Step 3: Determine critical points where numerator or denominator are zero: Numerator zero at $$x=6$$ Denominator zero at $$x=-6$$ (excluded from domain) Step 4: Test intervals divided by $$-6$$ and $$6$$: - For $$x < -6$$, denominator negative, numerator negative, fraction positive - For $$-6 < x < 6$$, denominator positive, numerator negative, fraction negative - For $$x > 6$$, denominator positive, numerator positive, fraction positive Step 5: Combine intervals where fraction $$\geq 0$$: $$(-\infty, -6) \cup [6, \infty)$$ with $$x \neq -6$$ (excluded) 2. Solve $$\frac{x+4}{x+3} \geq 0$$ Step 1: Critical points where numerator/denominator zero: Numerator zero at $$x = -4$$ Denominator zero at $$x = -3$$ (excluded) Step 2: Test intervals: - For $$x < -4$$, numerator negative, denominator negative, fraction positive - For $$-4 < x < -3$$, numerator positive, denominator negative, fraction negative - For $$x > -3$$, numerator positive, denominator positive, fraction positive Step 3: Combine intervals where fraction $$\geq 0$$: $$(-\infty, -4] \cup (-3, \infty)$$ 3. Solve $$\frac{x^2 - x - 12}{x+2} \geq 0$$ Step 1: Factor numerator: $$x^2 - x - 12 = (x - 4)(x + 3)$$ Step 2: Critical points: numerator zero at $$x=4, -3$$; denominator zero at $$x=-2$$ (excluded) Step 3: Write expression: $$\frac{(x-4)(x+3)}{x+2} \geq 0$$ Step 4: Test intervals divided by $$-3, -2, 4$$: - $$x < -3$$: numerator positive (since both factors negative times negative = positive), denominator negative, fraction negative - $$-3 < x < -2$$: numerator negative, denominator negative, fraction positive - $$-2 < x < 4$$: numerator negative, denominator positive, fraction negative - $$x > 4$$: numerator positive, denominator positive, fraction positive Step 5: Include points where numerator zero and denominator not zero: $$x=-3, 4$$ included Answer: $$[-3, -2) \cup [4, \infty)$$ 4. Solve $$\frac{x^2 -16}{x^2 -1} \leq 0$$ Step 1: Factor numerator and denominator: Numerator: $$x^2 -16 = (x-4)(x+4)$$ Denominator: $$x^2 -1 = (x-1)(x+1)$$ Step 2: Critical points: Numerator zero at $$-4, 4$$ Denominator zero at $$-1, 1$$ (excluded) Step 3: Expression: $$\frac{(x-4)(x+4)}{(x-1)(x+1)} \leq 0$$ Step 4: Intervals divided by $$-4, -1, 1, 4$$ - $$x < -4$$: Numerator positive, denominator positive, fraction positive - $$-4 < x < -1$$: Numerator negative, denominator positive, fraction negative - $$-1 < x < 1$$: Numerator negative, denominator negative, fraction positive - $$1 < x < 4$$: Numerator positive, denominator negative, fraction negative - $$x > 4$$: Numerator positive, denominator positive, fraction positive Step 5: Combine intervals where fraction $$\leq 0$$ plus zeros of numerator: $$(-4, -1) \cup (1,4)$$ with $$x \neq -1, 1$$ Include points where numerator zero: $$-4, 4$$ included Answer: $$[-4, -1) \cup (1, 4]$$ 5. Solve $$\frac{x+2}{x-4} \leq 1$$ Step 1: Move right side: $$\frac{x+2}{x-4} - 1 \leq 0$$ Step 2: Common denominator: $$\frac{x+2 - (x-4)}{x-4} \leq 0 \implies \frac{x+2 - x + 4}{x-4} \leq 0$$ Simplify numerator: $$\frac{6}{x-4} \leq 0$$ Step 3: Analyze fraction: Numerator is positive constant 6 Fraction $$\leq 0$$ when denominator negative: $$x-4 < 0 \Rightarrow x < 4$$ Step 4: Final solution: $$(-\infty, 4)$$ with $$x \neq 4$$ (domain restriction) Final answers: 1. $$(-\infty, -6) \cup [6, \infty)$$ 2. $$(-\infty, -4] \cup (-3, \infty)$$ 3. $$[-3, -2) \cup [4, \infty)$$ 4. $$[-4, -1) \cup (1, 4]$$ 5. $$(-\infty, 4)$$