SOLUTION: For a performance of a local play, 375 tickets were sold. The price of a ticket on the floor was 25$ and a balcony ticket went for 21$. If the total revenue was 8875$, how many of

Question 1098165: For a performance of a local play, 375 tickets were sold. The price of a ticket on the floor was 25$ and a balcony ticket went for 21$. If the total revenue was 8875$, how many of each kind of ticket were sold?
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52787)   (Show Source): You can put this solution on YOUR website!
.
1 - the system of 2 equations

F + B = 375, (1) (counting seats) 25*F + 21*B = 8875. (2) (counting money) To solve it, express F = 375 - B from (1) and substitute into (2). You will get a single equation for one unknown B: 25*(375 - B) + 21*B = 8875. (3) Simplify and solve for B. When you find B, calculate F from (1).

2 - One equation

Let B be the number of the Balcony seats. Then the number of the Floor seats is (375-B). The "money" equation is 25*(375 - B) + 21*B = 8875, exactly as (3).

If you want to see more solved problems of this type, look into the lesson
    - Using systems of equations to solve problems on tickets
in this site.

Also, you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lesson is the part of this online textbook under the topic "Systems of two linear equations in two unknowns".

Save the link to this online textbook together with its description

Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson

to your archive and use it when it is needed.

Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!

The response from tutor ikleyn is fine, although she doesn't take you all the way to the answer. But with her obvious start, I would definitely go a different direction than she does to get the answer.

She shows the two equations that come directly from the information given in the problem:

(1) (counting seats)
(2) (counting money)

But then she shows solving the system using substitution. When the two equations are both in this form (as they often are in this kind of problem), solving the system by elimination is (for me, at least) MUCH easier:

[multiplying the first equation by 25, to make the coefficients of F the same in the two equations]

125 balcony seats were sold, which means 250 floor seats.

Now here is precisely the way I solve this problem:

I first think "what would the total ticket cost be if all 375 seats were floor seats, and how does that compare with the actual total?" The answer is $25*375 = $9375, which is $500 more than the actual ticket sales amount.

Then I think "each balcony seat costs $4 less than a floor seat. Therefore the number of balcony seats is $500/$4 = 125."

If you look at those calculations, they are EXACTLY the calculations shown in my formal algebraic solution shown above.
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