1. Problem: Find the solution set of the equation $ \frac{x}{p(x)} - \frac{x}{r(x)} - 1 = 0 $ where $ p(x) = x - 4 $, $ r(x) = x^2 - 2x - 8 $. 2. Step 1: Write the equation explicitly: $$\frac{x}{x-4} - \frac{x}{x^2 - 2x - 8} - 1 = 0$$ 3. Step 2: Factorize the denominator $ r(x) $: $$x^2 - 2x - 8 = (x - 4)(x + 2)$$ 4. Step 3: Find common denominator $ (x-4)(x+2) $ and rewrite the equation: $$\frac{x(x+2)}{(x-4)(x+2)} - \frac{x}{(x-4)(x+2)} - 1 = 0$$ 5. Step 4: Combine the fractions: $$\frac{x(x+2) - x}{(x-4)(x+2)} - 1 = 0$$ Simplify numerator: $$x(x+2) - x = x^2 + 2x - x = x^2 + x$$ 6. Step 5: Write the equation as: $$\frac{x^2 + x}{(x-4)(x+2)} = 1$$ 7. Step 6: Multiply both sides by denominator: $$x^2 + x = (x-4)(x+2)$$ 8. Step 7: Expand right side: $$(x-4)(x+2) = x^2 + 2x - 4x - 8 = x^2 - 2x - 8$$ 9. Step 8: Set equation: $$x^2 + x = x^2 - 2x - 8$$ Subtract $ x^2 $ from both sides: $$x = -2x - 8$$ 10. Step 9: Solve for $ x $: $$x + 2x = -8$$ $$3x = -8$$ $$x = -\frac{8}{3}$$ 11. Step 10: Check for restrictions: denominators cannot be zero. $ x-4 \neq 0 \Rightarrow x \neq 4 $ $ x+2 \neq 0 \Rightarrow x \neq -2 $ Since $ x = -\frac{8}{3} $ is allowed, it is a valid solution. Final answer: The solution set is $ \left\{ -\frac{8}{3} \right\} $.