1. **Problem 1: Solve for $x$ and $y$ in equilateral triangle $\triangle ABC$ where $AB = BC$ and $AC = AB$.** Given: - $AB = 49$ - $BC = 3x + 1$ - $AC = 18y - 41$ Since $\triangle ABC$ is equilateral, all sides are equal: $$AB = BC = AC$$ So, $$49 = 3x + 1$$ $$49 = 18y - 41$$ 2. **Solve for $x$:** $$49 = 3x + 1$$ Subtract 1 from both sides: $$49 - 1 = 3x + 1 - 1$$ $$48 = 3x$$ Divide both sides by 3: $$\frac{\cancel{48}}{\cancel{3}} = \frac{3x}{3}$$ $$16 = x$$ 3. **Solve for $y$:** $$49 = 18y - 41$$ Add 41 to both sides: $$49 + 41 = 18y - 41 + 41$$ $$90 = 18y$$ Divide both sides by 18: $$\frac{\cancel{90}}{\cancel{18}} = \frac{18y}{18}$$ $$5 = y$$ 4. **Verify side lengths:** - $BC = 3x + 1 = 3(16) + 1 = 48 + 1 = 49$ - $AC = 18y - 41 = 18(5) - 41 = 90 - 41 = 49$ All sides equal 49, confirming the solution. **Final answer:** $$x = 16, \quad y = 5, \quad \text{each side} = 49$$