Question 1159406: Solving systems of equations: The football team sold 60 tickets to their fundraiser dinner. Student tickets cost $4 and adult tickets cost
$6. If they made $288, how many student and adult tickets were sold?
Answer by ikleyn(52897)   (Show Source):
You can put this solution on YOUR website! .
If x is the number of adult tickets sold, the the number of student tickets is (60-x).
The revenue equation is
4*(60-x) + 6x = 288.
Simplify and solve
240-4x + 6x = 288
2x = 288 - 240 = 48
x = 48/2 = 24.
ANSWER. 24 adult tickets, and the rest, 60-24 = 36 student tickets.
Solved.
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There are different methods of solving such problems.
Read the lessons
- Using systems of equations to solve problems on tickets
- Three methods for solving standard (typical) problems on tickets
in this site.
From these lessons, learn on how to solve such problems once and for all.
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.
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