1. **State the problem:** Solve the equation $$\frac{p}{x+r} + \frac{q}{x-r} = \frac{k}{2x}$$ for $x$. 2. **Formula and rules:** To solve rational equations, first find a common denominator to combine terms and eliminate fractions by multiplying both sides by the least common denominator (LCD). 3. **Find the LCD:** The denominators are $x+r$, $x-r$, and $2x$. The LCD is $$2x(x+r)(x-r).$$ 4. **Multiply both sides by the LCD:** $$2x(x+r)(x-r) \times \left( \frac{p}{x+r} + \frac{q}{x-r} \right) = 2x(x+r)(x-r) \times \frac{k}{2x}$$ 5. **Simplify each term:** - $$2x(x+r)(x-r) \times \frac{p}{x+r} = 2x p (x-r)$$ - $$2x(x+r)(x-r) \times \frac{q}{x-r} = 2x q (x+r)$$ - $$2x(x+r)(x-r) \times \frac{k}{2x} = k (x+r)(x-r)$$ 6. **Rewrite the equation:** $$2x p (x-r) + 2x q (x+r) = k (x+r)(x-r)$$ 7. **Expand terms:** - Left side: $$2x p x - 2x p r + 2x q x + 2x q r = 2x^2 p - 2x p r + 2x^2 q + 2x q r$$ - Right side: $$(x+r)(x-r) = x^2 - r^2$$ So right side is $$k (x^2 - r^2) = k x^2 - k r^2$$ 8. **Combine like terms on the left:** $$2x^2 p + 2x^2 q - 2x p r + 2x q r = 2x^2 (p+q) + 2x (q r - p r)$$ 9. **Set up the equation:** $$2x^2 (p+q) + 2x (q r - p r) = k x^2 - k r^2$$ 10. **Bring all terms to one side:** $$2x^2 (p+q) - k x^2 + 2x (q r - p r) + k r^2 = 0$$ 11. **Group terms:** $$x^2 (2(p+q) - k) + 2x (q r - p r) + k r^2 = 0$$ 12. **This is a quadratic in $x$:** $$A x^2 + B x + C = 0$$ where $$A = 2(p+q) - k$$ $$B = 2 (q r - p r)$$ $$C = k r^2$$ 13. **Use the quadratic formula:** $$x = \frac{-B \pm \sqrt{B^2 - 4 A C}}{2 A}$$ 14. **Final answer:** $$x = \frac{-2 (q r - p r) \pm \sqrt{(2 (q r - p r))^2 - 4 (2(p+q) - k) (k r^2)}}{2 (2(p+q) - k)}$$ This gives the solutions for $x$ in terms of $p$, $q$, $r$, and $k$.