A total of 78 seats for a local concert are sold producing a total revenue of $483. If seats cost either $2.50 or $10.50, how many of the $2.50 seats where sold?

Answer 1

Let \( x \) be the number of $2.50 seats sold, and \( y \) be the number of $10.50 seats sold. We can set up the following system of equations based on the problem statement:

1. The total number of seats sold: \[ x + y = 78 \]

2. The total revenue generated from the seats sold: \[ 2.50x + 10.50y = 483 \]

Now we can solve this system of equations.

First, we can solve the first equation for \( y \): \[ y = 78 - x \]

Next, substitute \( y \) into the second equation: \[ 2.50x + 10.50(78 - x) = 483 \]

Distributing \( 10.50 \): \[ 2.50x + 819 - 10.50x = 483 \]

Combining like terms: \[ -8.00x + 819 = 483 \]

Subtracting 819 from both sides: \[ -8.00x = 483 - 819 \] \[ -8.00x = -336 \]

Dividing both sides by -8: \[ x = \frac{-336}{-8} = 42 \]

Thus, the number of $2.50 seats sold is \( \boxed{42} \).

Now let's find \( y \) using the value of \( x \): \[ y = 78 - x = 78 - 42 = 36 \]

We can check our work by verifying the total revenue: \[ 2.50(42) + 10.50(36) = 105 + 378 = 483 \]

Since both equations are satisfied, the solution \( \boxed{42} \) for the number of $2.50 seats sold is correct.