jmc

We can split the expression $|x-3|=2x+4$ into two separate cases. In the first case, $$\begin{array}{rl}x-3& =2x+4\\ \Rightarrow -x& =7\\ \Rightarrow x& =-7\end{array}$$However if we plug this value of $x$ back into the original equation $|x-3|=2x+4$, we get that $|-7-3|=2(-7)+4$ or $10=-10$. Since this is clearly not a valid statement, the first case gives us no possible solutions.

In the second case, $$\begin{array}{rl}x-3& =-(2x+4)\\ x-3& =-2x-4\\ \Rightarrow 3x& =-1\\ \Rightarrow x& =-\frac{1}{3}.\end{array}$$If we plug $-\frac{1}{3}$ back into the initial equation, we get that $\left|-\frac{1}{3}-3\right|=2\left(-\frac{1}{3}\right)+4$ which simplifies to $\frac{10}{3}=\frac{10}{3}$. Since this is true, we can accept $x=-\frac{1}{3}$ as a valid solution to the equation. Therefore, the only value of $x$ that satisfies the given equation is $\boxed{-\frac{1}{3}}$.