\(\because DE \parallel BC\)
\(\therefore \frac{AB}{BD} = \frac{AC}{CE}\)
\[
\cfrac{2x}{x-3} = \cfrac{2x+3}{x-2}
\]
\[
2x(x-2) = (2x+3)(x-3)
\]
\[
2x^2 - 4x = 2x^2 - 6x + 3x - 9
\]
\[
2x^2 - 4x - 2x^2 + 3x = -9
\]
\[
-x = -9
\]
\[
x = 9
\]