1. **State the problem:** We need to find the cost of two types of concert tickets: a cheaper ticket and a more expensive ticket that costs 30 more than the cheaper one. A group buys 10 more cheaper tickets than expensive tickets, and the total cost is 1800. 2. **Define variables:** Let $x$ be the cost of the cheaper ticket. Let $y$ be the cost of the more expensive ticket. Let $n$ be the number of expensive tickets bought. 3. **Write equations from the problem:** - The more expensive ticket costs 30 more than the cheaper one: $$y = x + 30$$ - The group buys 10 more cheaper tickets than expensive tickets: Number of cheaper tickets = $n + 10$ - Total cost is 1800: $$x(n + 10) + y n = 1800$$ 4. **Substitute $y$ from the first equation into the total cost equation:** $$x(n + 10) + (x + 30) n = 1800$$ 5. **Expand and simplify:** $$x n + 10 x + x n + 30 n = 1800$$ $$2 x n + 10 x + 30 n = 1800$$ 6. **We have two variables $x$ and $n$ but one equation. We need to express $x$ in terms of $n$ or vice versa. Rearrange:** $$2 x n + 10 x = 1800 - 30 n$$ $$x(2 n + 10) = 1800 - 30 n$$ $$x = \frac{1800 - 30 n}{2 n + 10}$$ 7. **Since $x$ must be positive and reasonable, try integer values of $n$ to find integer $x$ and $y$: try $n=10$:** $$x = \frac{1800 - 30 \times 10}{2 \times 10 + 10} = \frac{1800 - 300}{20 + 10} = \frac{1500}{30} = 50$$ 8. **Calculate $y$:** $$y = x + 30 = 50 + 30 = 80$$ 9. **Check total cost:** Number of cheaper tickets = $n + 10 = 10 + 10 = 20$ Total cost = $50 \times 20 + 80 \times 10 = 1000 + 800 = 1800$ 10. **Answer:** The cheaper ticket costs 50 and the more expensive ticket costs 80.