Let's define two variables to represent the number of tickets sold:

• Let $x$ be the number of student tickets sold.
• Let $y$ be the number of adult tickets sold.

From the problem, we have two pieces of information:

1. The total number of tickets sold was 200. This gives us the equation: $x + y = 200$

2. The total revenue from the tickets was $1,250. Since student tickets cost $5 and adult tickets cost $8, this gives us the second equation: $5x + 8y = 1250$

Now, we can solve this system of equations.

Step 1: Solve the first equation for one variable.

From $x + y = 200$, we can solve for $x$: $x = 200 - y$

Step 2: Substitute into the second equation.

Substitute $x = 200 - y$ into $5x + 8y = 1250$: $5(200 - y) + 8y = 1250$

Step 3: Simplify and solve for $y$.

Expand the equation: $1000 - 5y + 8y = 1250$ Combine like terms: $1000 + 3y = 1250$ Subtract 1000 from both sides: $3y = 250$ Solve for $y$: $y = \frac{250}{3} \approx 83.33$

So, the number of adult tickets sold is approximately 83.33 tickets, which is not a feasible solution. Therefore, a valid solution cannot be given unless the original problem uses rounding or other approximations.