Question 1132796: . Adult tickets cost 5 dollars and child tickets cost 2 dollars. The box office recorded $190 in ticket sales. A total of 50 tickets were sold. How many of those tickets were child tickets?
Write a system and solve to answer the question. One of your equations is x + y = 50.
Found 4 solutions by MathLover1, ikleyn, greenestamps, MathTherapy:
Answer by MathLover1(20849)   (Show Source):
Answer by ikleyn(52781)  (Show Source):
You can put this solution on YOUR website! .
Your first equation is
x + y = 50 (counting the tickets)
Your second equation is your "money" equation, or "revenue"
5x + 2y = 190 (dollars)
So, you have the system of 2 equations in 2 unknowns
x + y = 50, (1) and
5x + 2y = 190. (2)
There are different methods to solve the system.
Since you are a beginner, I will show you the simplest method, which is the Substitution method.
From equation (1), express y = 50-x and substitute it into the second equation, replacing y there. You will get
5x + 2*(50-x) = 190.
Simplify it step by step and solve for x.
5x + 100 - 2x = 190.
5x - 2x = 190 - 100
3x = 90 ====> x = 90/3 = 30.
We just found x, or the number of adults. It is 30.
Now substitute the found value of x= 30 into equation (1).
30 + y = 50
y = 50-30 = 20
So, the remained number of children is 20.
ANSWER. 30 adults and 20 children.
CHECK. 30*5 + 20*2 = 150 + 40 = 190 dollars is your REVENUE, exactly as the problem states.
Hence, the found values are the correct solution !
Solved.
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It is a standard and typical ticket problem.
For ticket problems, see 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".
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
You have two pieces of information which you can use to write equations involving the numbers of adult (x) and child (y) tickets:
(1) x+y = 50 the total number of tickets was 50
(2) 5x+2y = 190 the total cost of the tickets -- $5 per adult and $2 per child -- was 190
You have two responses, in both of which the tutors solved the system of equations by substitution -- solving one equation for one of the variables and substituting the expression for that variable in the other equation.
That is one of the two most common methods for solving a system of two equations. One of the tutors even says that substitution is the easiest method.
For me, when both of the equations are in the form they are for this problem, the easiest method is elimination.
With the method of elimination, you modify the given equations in such a way that either adding them or subtracting one from the other eliminates one of the variables.
So my chosen path for solving the system of equations for this problem would be this:
double equation (1): x+y = 50 --> 2x+2y = 100
subtract this new equation from the second given equation:
(5x+2y)-(2x+2y) = 190-100
3x = 90
x = 30
Then x=30 and x+y=50 --> y = 20
ANSWER: 20 adults and 30 children
Note you should understand both methods for solving systems of equations. Which one is easiest will depend on the individual student.
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Again tutor @MathTherapy has proclaimed that he is better than anyone else, by saying that substitution is much easier than elimination for this problem, and that you should not let anyone tell you otherwise.
Ignore his arrogant and absurd proclamations. Learn both methods and use whichever one suits you for a particular problem.
Answer by MathTherapy(10552)  (Show Source):
You can put this solution on YOUR website!
. Adult tickets cost 5 dollars and child tickets cost 2 dollars. The box office recorded $190 in ticket sales. A total of 50 tickets were sold. How many of those tickets were child tickets?
Write a system and solve to answer the question. One of your equations is x + y = 50.
Since you want one of the equations to be x + y = 50, then let the number of children's tickets be x, which makes the number of adults' tickets, y
Then x + y = 50 becomes: y = 50 - x ----- eq (i)
Then other equation is: 2x + 5y = 190 --- eq (ii)
5(50 - x) + 2x = 190 ------ Substituting 50 - x for y
250 - 5x + 2x = 190
- 5x + 2x = 190 - 250
- 3x = - 60
x, or number of children's tickets = 
SUBSTITUTION, in this case is much easier than ELIMINATION. Let no-one tell you otherwise!
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