Linda C. answered • 12/30/14
Tutor
5
(1)
Engaging teacher fro Calculus, Precalculus, Trigonometry, and Algebra
A = adult ticket
C = children's ticket
2A+5C=$48
2A+3C=$32
Use elimination, since both equations have 2A...you can subtract the second from the first to get:
2C=16
C=8
Plug C=8 into either equation from here. I'll use the first:
2A+5(8)=48
2A+40=48
2A=8
A=4
So a child's ticket is $8 and an adult's ticket is $4
C = children's ticket
2A+5C=$48
2A+3C=$32
Use elimination, since both equations have 2A...you can subtract the second from the first to get:
2C=16
C=8
Plug C=8 into either equation from here. I'll use the first:
2A+5(8)=48
2A+40=48
2A=8
A=4
So a child's ticket is $8 and an adult's ticket is $4
Mary Ann F.
It does if the event is something especially meant for children. But it is hard to get young children to come to an event without their parents!
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12/30/14
Linda C.
tutor
I was initially thrown by that too, but I like Mary Ann's point. The event is obviously something geared toward children, hence the higher cost for them. The math is correct.
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12/30/14
Luke P.
12/30/14