Answer:
a) 47
b) 61
Step-by-step explanation: We can setup two equations from the information given above:
2c + 4b = 14 where c is the cost if one carousel ticket and b is the cost of four bumper car tickets.
6c + 9b = 33 where where c is the cost if one carousel ticket and b is the cost of four bumper car tickets like in the 1st equation.
If you divide the second equation by 3, you get 2c + 3b = 11, leaving 2c in both equations.
If you subtract both equations by the elimination method:
2c + 4b = 14
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2c + 3b = 11
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b = 3, which solves for the cost of one bumper car ticket
If you plug this value back into the first (or second) equation,
ex. 2c + 4(3) = 14 -> 2c = 2 -> c=1
You get c = 1, which solves for the cost of one carousel ticket.
Then you plug these values (c=1 and b=3) in a) and b):
a) 8*1 + 13*3 = 47
b) 10*1 + 17*3 = 61