1. **State the problem:** We have two equations representing ticket purchases: $$2x + 5y = 64$$ $$4x + 3y = 72$$ where $x$ is the price of an adult ticket and $y$ is the price of a children's ticket. 2. **Goal:** Solve the system using elimination to find the prices of adult and children's tickets. 3. **Elimination method:** We want to eliminate one variable by making the coefficients of $x$ or $y$ the same in both equations. 4. Multiply the first equation by 2 to match the coefficient of $x$ in the second equation: $$2(2x + 5y) = 2(64) \Rightarrow 4x + 10y = 128$$ 5. Now subtract the second equation from this new equation: $$ (4x + 10y) - (4x + 3y) = 128 - 72 $$ $$4x - 4x + 10y - 3y = 56$$ $$7y = 56$$ 6. Solve for $y$: $$y = \frac{56}{7} = 8$$ 7. Substitute $y=8$ back into the first original equation: $$2x + 5(8) = 64$$ $$2x + 40 = 64$$ $$2x = 64 - 40 = 24$$ $$x = \frac{24}{2} = 12$$ 8. **Interpretation:** The solution $(x, y) = (12, 8)$ means the adult ticket costs 12 and the children's ticket costs 8. 9. **Summary:** - Adult ticket price: 12 - Children's ticket price: 8 This satisfies both purchase scenarios given.