1. **Problem Statement:** Given rhombus ABCD with AE = 3x - 5 and CE = x + 3, find $x$, $AE$, $CE$, and $AC$. 2. **Formula and Properties:** In a rhombus, the diagonals bisect each other. Therefore, $AE = CE$ because E is the midpoint of diagonal AC. 3. **Set up the equation:** Since $AE = CE$, we have: $$3x - 5 = x + 3$$ 4. **Solve for $x$:** $$3x - x = 3 + 5$$ $$2x = 8$$ $$x = 4$$ 5. **Find $AE$ and $CE$ by substituting $x=4$:** $$AE = 3(4) - 5 = 12 - 5 = 7$$ $$CE = 4 + 3 = 7$$ 6. **Find $AC$:** Since $E$ is midpoint of $AC$, $$AC = AE + CE = 7 + 7 = 14$$ **Final answers:** $$x = 4, AE = 7, CE = 7, AC = 14$$