1. **State the problem:** We need to solve the equation $$\frac{2x+4}{x-3} = 3$$ for $x$. 2. **Recall the formula and rules:** To solve a rational equation like this, multiply both sides by the denominator to eliminate the fraction, but remember $x \neq 3$ because the denominator cannot be zero. 3. **Multiply both sides by $x-3$:** $$\cancel{\frac{2x+4}{x-3}} \times (x-3) = 3 \times (x-3)$$ which simplifies to $$2x + 4 = 3(x - 3)$$ 4. **Expand the right side:** $$2x + 4 = 3x - 9$$ 5. **Bring all terms to one side to isolate $x$: subtract $2x$ from both sides:** $$\cancel{2x} + 4 = 3x - 9 - \cancel{2x}$$ which simplifies to $$4 = x - 9$$ 6. **Add 9 to both sides to solve for $x$:** $$4 + 9 = x - 9 + 9$$ $$13 = x$$ 7. **Check the solution:** Substitute $x=13$ back into the denominator $x-3$: $$13 - 3 = 10 \neq 0$$ So $x=13$ is valid. **Final answer:** $$x = 13$$