Question 270719: The admission fee for Knoebels amusement park is $5 for children and $9 for adults. On a certain day, 450 people enter the park and the admission fees collected totaled $3,050. How many children and how many adults were admitted?
(Here's what I tried):
Variables:
c= Child admission
a= Adult admission
c+a=450
$5c+$9a=$3,050
changed c+a=450 to c=450-a
substitution:
$5(450-a)+$9a=$3,050
Not sure if my equations are right. Based off of how I solved a similiar problem. And I don't know how to simplify from there.
Could you please tell me if I'm on the right track and if so, how to solve from where I stopped?
Answer by dabanfield(803)   (Show Source):
You can put this solution on YOUR website! Great progress so far:
c= Child admission
a= Adult admission
c+a=450
$5c+$9a=$3,050
changed c+a=450 to c=450-a
substitution:
$5(450-a)+$9a=$3,050
Not sure if my equations are right. Based off of how I solved a similiar problem. And I don't know how to simplify from there.
Could you please tell me if I'm on the right track and if so, how to solve from where I stopped?
So starting from where you left off:
5(450-a) + 9a = 3050
2250 - 5a + 9a = 3050
9a - 5a = 3050 - 2250
4a = 800
a = 200
c = 450 - a
c = 450 - 200
c = 250
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