math problem to long so in description

Question

A 1300-seat theater sells two types of tickets for a concert. Premium seats sell for $30 each and regular seats sell for $20 each. At one event $31,830 was collected in ticket sales with 30 seats left unsold. How many of each type of ticket was sold?

Answer 1

P+ R = 1300-30 = 1270

30P + 20R = 31830

20P + 20R = 20(1270) = 25400

subtract to get

10P = 6430

P = 643 premium

R=1270-643 = 627 regular

Answer 2

Hi Cole,

To solve this you need to create a system of equations - or 2 equations where they eliminate one variable.

For this problem I'll use "P" for premium and "R" for regular. We know 30 tickets weren't sold so that brings the total to 1,270. Therefore:

P + R = 1,270

The next equation is your pricing. You know that P sold for $30 each while R sold for $20 each and the final sales was $31,830. With that info you can set up your next equation:

30P + 20R = 31,830

Now let's put the equations together

30P + 20R = 31,830

P + R = 1,270

Nothing to eliminate off the bat, however let's multiply the bottom by -30. This will allow us to eliminate the "P". Be sure to multiply all numbers in that specific equation by -30. You should get -30P + -30R = -38,100

30P + 20R = 31,830

-30P - 30R = -38,100

-----------------------------

-10R = -6,270

Divide each side by -10, you'll get R = 627. Use that in your original P + R = 1,270 and you'll find the "P".

Pro tip: You should check your answer to make sure it work in the other equation too 30P + 20R = $31,830.

I hope this helped! If you need any other assistance, please don't hesitate to reach out - I tutor through video conference. Best of luck!