Question 193145: Hi..
My question is ..
In one day a ticket agent sold 395 tickets to the circus. A child ticket cost $22 and an adult ticket cost $28. The total amount collected that day was $10,130. How many of each type of ticket did the agent sell that day?
well I did..
22 c+28 a=10130
c+a=395 ___ but i changed it into..c=395-9
I dont know if this is right
But i was trying use the substitution method ..
will you help me..
to find the answer
Answer by Earlsdon(6294)   (Show Source):
You can put this solution on YOUR website! You are definitely on the right track!
Let c = the number of child tickets and a = the number of adult tickets. (It's always appropriate to define your variables).
As you have correctly stated:
22(c)+28(a) = 10,130
a+c = 395 which can be expressed as c = 395-a Now you can substitute c = 395-a into the first equation, to get:
22(395-a)+28a = 10,130 Now you can solve for a.
8690-22a + 28a = 10,130 Simplify.
8690 + 6a = 10,130 Subtract 8690 from both sides.
6a = 1440 Divide both sides by 6.
a = 240 This is the number of adult tickets sold.
c = 395-a
c = 395-240
c = 155 This is the number of child tickets sold.
Check:
c+a = 395
155 + 240 = 395 Total number of tickets sold.
$22(c)+$28(a) = $10,130
$22(155) + $28(240) = $10,130
$3410 + $6720 = $10,130
$10,130 = $10,130
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