1. The problem is to solve the equation $$\frac{2x+4}{x-3} = 3$$ for $x$. 2. We start by multiplying both sides of the equation by the denominator $x-3$ to eliminate the fraction: $$\cancel{(x-3)} \cdot \frac{2x+4}{\cancel{x-3}} = 3 \cdot (x-3)$$ which simplifies to $$2x + 4 = 3(x - 3)$$ 3. Next, expand the right side: $$2x + 4 = 3x - 9$$ 4. Now, bring all terms involving $x$ to one side and constants to the other: $$2x - 3x = -9 - 4$$ which simplifies to $$-x = -13$$ 5. Multiply both sides by $-1$ to solve for $x$: $$x = 13$$ 6. Finally, check that $x=13$ does not make the denominator zero. Since $13 - 3 = 10 \neq 0$, the solution is valid. The solution to the equation is $$\boxed{13}$$.