SOLUTION: Solve the following problem using two variables and a system of two equations. Solve the system by the method of your choice. The City Zoo has different admission prices for adult

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Solve the following problem using two variables and a system of two equations. Solve the system by the method of your choice. The City Zoo has different admission prices for adult      Log On


   

Question 1148837: Solve the following problem using two variables and a system of two equations. Solve the system by the method of your choice.
The City Zoo has different admission prices for adults and children. When three adults and two children went to the​ zoo, the bill was​ $78.05. If two adults and three children got in for ​$73.90​, then what is the price of an​ adult's ticket and what is the price of a​ child's​ ticket?
Found 2 solutions by ikleyn, Theo:
Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.
It is a standard tickets problem.

There are different methods of solving such problems.

In this site,  there is the lesson
    - Using systems of equations to solve problems on tickets
explaining and showing all basic methods of solving such problems.

From this lesson,  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 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 Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
x = the price of an adult ticket.
y = the price of a child ticket.

you have two equations that need to be solved simultaneously.
they are:
3x + 2y = 78.05
2x + 3y = 73.90

multiply both sides of the first equation by 2 and multiply both sides of the second equation to 3 to get:
6X + 4Y = 156.10
6X + 9Y = 221.70
subtract the first equation from the second to get:
5y = 65.60
solve for y to get:
y = 13.12
replace y in the first original equation of 3x + 2y = 78.05 and solve for x to get:
x = 17.27

you have:
x = 17.27
y = 13.12

go back to the original two equations to get:
3x + 2y = 78.05 becomes 51.81 + 26.24 = 78.05 which becomes 78.05 = 78.05, which is true.
2x + 3y = 73.90 becomes 34.54 + 39.36 = 73.90 which becomes 73.90 = 73.90, which is true.

both original equations are true when x = 17.27 and y = 13.13.
this confirms those values are good.

your solution is that the price of an adult ticket is 17.27 and the price of a child ticket is 13.12.