1. **State the problem:** Solve the rational equation $$\frac{9}{x^2 - 6x + 9} = \frac{3x}{x^2 - 3x}$$ and state any restrictions on $x$. 2. **Identify restrictions:** The denominators cannot be zero. - For $x^2 - 6x + 9 = (x-3)^2$, so $x \neq 3$. - For $x^2 - 3x = x(x-3)$, so $x \neq 0$ and $x \neq 3$. 3. **Rewrite the equation:** $$\frac{9}{(x-3)^2} = \frac{3x}{x(x-3)}$$ Simplify the right side denominator: $$\frac{3x}{x(x-3)} = \frac{3x}{x(x-3)}$$ (no simplification yet). 4. **Cross-multiply:** $$9 \cdot x(x-3) = 3x \cdot (x-3)^2$$ 5. **Expand both sides:** Left: $$9x(x-3) = 9x^2 - 27x$$ Right: $$3x(x-3)^2 = 3x(x^2 - 6x + 9) = 3x^3 - 18x^2 + 27x$$ 6. **Set equation:** $$9x^2 - 27x = 3x^3 - 18x^2 + 27x$$ 7. **Bring all terms to one side:** $$0 = 3x^3 - 18x^2 + 27x - 9x^2 + 27x$$ Simplify: $$0 = 3x^3 - 27x^2 + 54x$$ 8. **Factor out common terms:** $$0 = 3x(x^2 - 9x + 18)$$ 9. **Solve each factor:** - $3x = 0 \Rightarrow x = 0$ (excluded from restrictions) - Solve quadratic $x^2 - 9x + 18 = 0$ 10. **Quadratic formula:** $$x = \frac{9 \pm \sqrt{81 - 72}}{2} = \frac{9 \pm \sqrt{9}}{2} = \frac{9 \pm 3}{2}$$ 11. **Solutions:** - $x = \frac{9 + 3}{2} = 6$ - $x = \frac{9 - 3}{2} = 3$ (excluded from restrictions) 12. **Final valid solution:** $$x = 6$$ --- 1. **State the problem:** Solve the rational equation $$\frac{1}{x^2} + \frac{x+3}{2x} = \frac{1}{2}$$ and state any restrictions on $x$. 2. **Identify restrictions:** - $x^2 \neq 0 \Rightarrow x \neq 0$ - $2x \neq 0 \Rightarrow x \neq 0$ 3. **Find common denominator:** The common denominator is $2x^2$. 4. **Rewrite each term with common denominator:** $$\frac{1}{x^2} = \frac{2}{2x^2}$$ $$\frac{x+3}{2x} = \frac{(x+3)x}{2x^2} = \frac{x^2 + 3x}{2x^2}$$ $$\frac{1}{2} = \frac{x^2}{2x^2}$$ 5. **Rewrite equation:** $$\frac{2}{2x^2} + \frac{x^2 + 3x}{2x^2} = \frac{x^2}{2x^2}$$ 6. **Combine left side:** $$\frac{2 + x^2 + 3x}{2x^2} = \frac{x^2}{2x^2}$$ 7. **Cross-multiply:** $$2 + x^2 + 3x = x^2$$ 8. **Simplify:** $$2 + 3x = 0$$ 9. **Solve for $x$:** $$3x = -2 \Rightarrow x = -\frac{2}{3}$$ 10. **Check restrictions:** $x = -\frac{2}{3} \neq 0$, so valid. --- **Final answers:** - For the first equation: $x = 6$ - For the second equation: $x = -\frac{2}{3}$ **Restrictions:** - $x \neq 0, 3$ for the first equation - $x \neq 0$ for the second equation