1. **State the problem:** Solve the equation $$\frac{2x+4}{x} + \frac{x+2}{x-2} = \frac{3x}{x-2}$$ for $x$. 2. **Identify the domain restrictions:** The denominators cannot be zero, so $x \neq 0$ and $x \neq 2$. 3. **Find a common denominator:** The denominators are $x$ and $x-2$, so the common denominator is $x(x-2)$. 4. **Multiply both sides of the equation by the common denominator to clear fractions:** $$x(x-2) \times \left(\frac{2x+4}{x} + \frac{x+2}{x-2}\right) = x(x-2) \times \frac{3x}{x-2}$$ 5. **Simplify each term:** $$ (x-2)(2x+4) + x(x+2) = 3x^2 $$ 6. **Expand the terms:** $$ (x-2)(2x+4) = 2x^2 + 4x - 4x - 8 = 2x^2 - 8 $$ $$ x(x+2) = x^2 + 2x $$ 7. **Rewrite the equation:** $$ 2x^2 - 8 + x^2 + 2x = 3x^2 $$ 8. **Combine like terms on the left:** $$ 3x^2 + 2x - 8 = 3x^2 $$ 9. **Subtract $3x^2$ from both sides:** $$ 3x^2 + 2x - 8 - \cancel{3x^2} = \cancel{3x^2} $$ $$ 2x - 8 = 0 $$ 10. **Solve for $x$:** $$ 2x = 8 $$ $$ x = \frac{8}{2} $$ $$ x = 4 $$ 11. **Check domain restrictions:** $x=4$ is allowed since it is not 0 or 2. **Final answer:** $$x = 4$$