Drawing a diagram I saw that EH = FG so 2x=x+3, and x=3.
I felt this was a bit too simple, though, so is it correct?
Your instinct was correct: EG isn't necessarily equal to FG.
Here's a solution:
Notice that DGH and DFE are similar. We can set up an equation:
\(\frac{x-4}{x-7}=\frac{(x-4)+2x}{(x-7)+(x+3)}\\ \frac{x-4}{x-7}=\frac{3x-4}{2x-4}\\ (x-4)(2x-4)=(x-7)(3x-4)\\ 2x^2-12x+16=3x^2-25x+28\\ 0=x^2-13x+12\\ (x-12)(x-1)=0\\ x=1,x=12\)
1 is extraneous because it will be negative for x-4, which is impossible.
Therefore the answer is \(\boxed{24}\)