1. **State the problem:** Solve the equation $$\frac{2}{x} + \frac{3}{x-3} = 1$$ for $x$. 2. **Find a common denominator:** The denominators are $x$ and $x-3$. The common denominator is $x(x-3)$. 3. **Rewrite each term with the common denominator:** $$\frac{2}{x} = \frac{2(x-3)}{x(x-3)}$$ $$\frac{3}{x-3} = \frac{3x}{x(x-3)}$$ 4. **Combine the fractions:** $$\frac{2(x-3) + 3x}{x(x-3)} = 1$$ 5. **Multiply both sides by $x(x-3)$ to clear the denominator:** $$2(x-3) + 3x = x(x-3)$$ 6. **Expand both sides:** $$2x - 6 + 3x = x^2 - 3x$$ 7. **Combine like terms on the left:** $$5x - 6 = x^2 - 3x$$ 8. **Bring all terms to one side to form a quadratic equation:** $$0 = x^2 - 3x - 5x + 6$$ $$0 = x^2 - 8x + 6$$ 9. **Use the quadratic formula to solve for $x$:** The quadratic formula is $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=-8$, and $c=6$. 10. **Calculate the discriminant:** $$b^2 - 4ac = (-8)^2 - 4(1)(6) = 64 - 24 = 40$$ 11. **Find the roots:** $$x = \frac{8 \pm \sqrt{40}}{2} = \frac{8 \pm 2\sqrt{10}}{2} = 4 \pm \sqrt{10}$$ 12. **Check for restrictions:** The denominators $x$ and $x-3$ cannot be zero, so $x \neq 0$ and $x \neq 3$. 13. **Final answer:** $$x = 4 + \sqrt{10}, 4 - \sqrt{10}$$ Both values are valid since neither equals 0 nor 3.