1. **State the problem:** Solve the equation $$\frac{2}{x^2 - 4x + 3} = \frac{2x}{x - 1} - \frac{x}{x - 3}$$ for $x$. 2. **Factor the quadratic denominator:** Note that $$x^2 - 4x + 3 = (x - 1)(x - 3)$$. 3. **Rewrite the equation using the factored form:** $$\frac{2}{(x - 1)(x - 3)} = \frac{2x}{x - 1} - \frac{x}{x - 3}$$ 4. **Find a common denominator on the right side:** The common denominator is $(x - 1)(x - 3)$. 5. **Rewrite the right side with the common denominator:** $$\frac{2x(x - 3)}{(x - 1)(x - 3)} - \frac{x(x - 1)}{(x - 3)(x - 1)} = \frac{2x(x - 3) - x(x - 1)}{(x - 1)(x - 3)}$$ 6. **Simplify the numerator on the right side:** $$2x(x - 3) - x(x - 1) = 2x^2 - 6x - x^2 + x = (2x^2 - x^2) + (-6x + x) = x^2 - 5x$$ 7. **Rewrite the equation:** $$\frac{2}{(x - 1)(x - 3)} = \frac{x^2 - 5x}{(x - 1)(x - 3)}$$ 8. **Since denominators are equal and not zero (exclude $x=1$ and $x=3$), set numerators equal:** $$2 = x^2 - 5x$$ 9. **Bring all terms to one side:** $$x^2 - 5x - 2 = 0$$ 10. **Use the quadratic formula:** $$x = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1} = \frac{5 \pm \sqrt{25 + 8}}{2} = \frac{5 \pm \sqrt{33}}{2}$$ 11. **Check for restrictions:** $x \neq 1$ and $x \neq 3$, neither solution equals these values. **Final answer:** $$x = \frac{5 + \sqrt{33}}{2}, \frac{5 - \sqrt{33}}{2}$$