SOLUTION: A play was attended by 320 people. Regular tickets were $15 each, while student tickets were priced at $10. The total receipts were $4,200. Set up a system of linear equations in

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Question 1155419: A play was attended by 320 people. Regular tickets were $15 each, while student tickets were priced at $10. The total receipts were $4,200. Set up a system of linear equations in order to determine how many of each types of tickets were sold
Found 4 solutions by MathLover1, ikleyn, greenestamps, MathTherapy:
Answer by MathLover1(20850)   (Show Source): You can put this solution on YOUR website!

A play was attended by people.
Regular tickets were $ each, while student tickets were priced at $.
total number of tickets:
.....eq.1
The total receipts were $.
$.......eq.2
Set up a system of linear equations in order to determine how many of each types of tickets were sold.
.....eq.1......both sides multiply by
.......eq.2
------------------------------------------
.....eq.1
.......eq.2
------------------------------------------subtract eq.2 from eq.1






go to
.....eq.1, substitute



of regular tickets and of student tickets were sold

Answer by ikleyn(52798)   (Show Source): You can put this solution on YOUR website!
.

The setup is this system of 2 equations in 2 unknown


      x +   y =  320     (1)    (counting tickets)

    15x + 10y = 4200     (2)    (counting money),


where x is the number of 15-dollar (regular) tickets, and y is the number of 10-dollar (student) tickets.

Usually, for such simple problems, the setup, i.e. writing the system of equations,
is considered as absolutely trivial (routine) step.

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

If you want (if you need) to see how to setup typical problems with 2x2 system of equations, look into the lessons
    - Oranges and grapefruits
    - Using systems of equations to solve problems on tickets
    - Three methods for solving standard (typical) problems on tickets
    - Using systems of equations to solve problems on shares
    - Using systems of equations to solve problems on investment
    - Two mechanics work on a car
    - The Robinson family and the Sanders family each used their sprinklers last summer
    - Roses and vilolets
    - Counting calories and grams of fat in combined food
    - A theater group made appearances in two cities
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.


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


Here are two (very similar) quick and easy non-algebraic methods for solving the problem (if an algebraic solution is not required!).

Method 1....

(1) 320 tickets all at $10 would bring in $3200; the actual total was $4200, which is $1000 greater.
(2) Each regular ticket costs $5 more than a student ticket. The number of regular tickets sold, in order to make that additional $1000, is 1000/5 = 200.

ANSWER: 200 regular tickets; 120 student tickets.

CHECK: 200(15)+120(10) = 3000+1200 = 4200

Method 2....

(1) 320 tickets at $10 would bring in $3200; 320 all at $15 would bring in $4800.
(2) The actual total of $4200 is 1000/1600 = 5/8 of the way from $3200 to $4800, so 5/8 of the tickets were the more expensive tickets.

ANSWER: 5/8 of the 320 tickets, or 200, were the $15 tickets; the other 120 were the less expensive.
Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!

A play was attended by 320 people. Regular tickets were $15 each, while student tickets were priced at $10. The total receipts were $4,200. Set up a system of linear equations in order to determine how many of each types of tickets were sold
Let regular and student tickets be r and s, respectively

Then we get: r + s = 320___r = 320 - s ------- eq (i)

Also, 15r + 10s = 4,200_____5(3r + 2s) = 5(840)______3r + 2s = 840 ------ eq (ii)

3(320 - s) + 2s = 840 ------ Substituting 320 - s for r in eq (ii)

960 - 3s + 2s = 840

- s = 840 - 960

- s = - 120

s, or 

You should be able to find the number of regular tickets sold! 

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