SOLUTION: I'm not sure how to go about solving this problem. There are 2 dollar amounts for the students. Can you please show me how to set up this equation? Thanks in advance Ticket

Algebra ->  Equations -> SOLUTION: I'm not sure how to go about solving this problem. There are 2 dollar amounts for the students. Can you please show me how to set up this equation? Thanks in advance Ticket      Log On


   



Question 1137316: I'm not sure how to go about solving this problem. There are 2 dollar amounts for the students.
Can you please show me how to set up this equation?
Thanks in advance
Tickets for the spring musical at Bayview High School were being sold to students for $8 in advance of the performance, $10 on the day of the performance, and $12 for adults (no matter when the ticket was purchased). The financial report after the musical performances showed that 1250 tickets were sold with receipts totaling $13,400. The number of adult tickets sold exceeded the total number of student tickets sold by 150. How many tickets of each type were sold for the musical?

Found 2 solutions by josmiceli, ikleyn:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +a+ = number of $8 tickets sold
Let +b+ = number of $10 tickets sold
Let +c+ = number of $12 tickets sold
-----------------------------------------------
(1) +a+%2B+b+%2B+c+=+1250+
(2) +c+=+a+%2B+b+%2B+150+
(3) +8a+%2B+10b+%2B+12c+=+13400+
-----------------------------------------
There are 3 equations with 3 unknowns, so it's solvable

Answer by ikleyn(52886) About Me  (Show Source):
You can put this solution on YOUR website!
.

            Tutor @josmicely showed you how to setup the problem using 3 unknown and the system of 3 equations.

            In my post I will show you how you can easily to setup the problem using ONLY 2 unknowns and 2 equations.


Let x be the number of the advance student tickets and

let y be the number of the same day student tickets.


Then the number of the adults tickets is  (x + y + 150), according to the condition.


Then you can write this system of 2 equations, based on the condition


     x +   y + (x + y + 150) =  1250     (1)   (counting all tickets)

    8x + 10y + 12*(x+y+150)  = 13400     (2)   (counting money)


Simplify the equations to get


    2x + 2y = 1250 - 150,                 (1')

    20x + 22y = 13400 - 12*150,           (2')


or, equivalently,


    2x  +  2y =  1100,

    20x + 22y = 11600.


You can do even one more step simplifying


      x  + y =   550,        (3)

     10x + 11y = 5800.       (4)


Should I explain that 2x2-system of linear equations is much easier to solve than 3x3-system of linear equations ?

My setup is completed, and I think that it is your goal to learn on how to do it.

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

To see many other similar solved problem, look into the lesson
    - HOW TO algebreze and solve this problem on 2 equations in 2 unknowns
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 lessons are 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.