1. **Problem:** Find $x$ in triangle ACB where segment GH is parallel to CB. Given: $AG=10$, $GH=2x+6$, $CB=25$, $CH=55$. 2. **Formula:** When a segment is parallel to one side of a triangle, corresponding sides are proportional. 3. **Set up proportion:** $$\frac{AG}{GH} = \frac{CH}{CB}$$ 4. **Substitute values:** $$\frac{10}{2x+6} = \frac{55}{25}$$ 5. **Cross multiply:** $$10 \times 25 = 55 \times (2x+6)$$ 6. **Simplify:** $$250 = 55(2x+6) = 110x + 330$$ 7. **Solve for $x$:** $$110x = 250 - 330 = -80$$ $$x = \frac{-80}{110} = -\frac{8}{11}$$ **Answer:** $x = -\frac{8}{11}$