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Write a system of equations to solve the following problem. Let c be the number of child tickets and a be the number of adult tickets.
Each child ticket for a ride costs $3, while each adult ticket costs $5. If the ride collected a total of $115, and 33 tickets were sold, how many of each type of ticket were sold?



27 small and 6 large

20 small and 10 large

10 small and 20 large

8 small and 22 large
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This next question confuses me...sorry
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Write a system of equations to solve the following problem. Let c be the number of child tickets and a be the number of adult tickets.
Each child ticket for a ride costs $3, while each adult ticket costs $5. If the ride collected a total of $115, and 33 tickets were sold, how many of each type of ticket were sold?
(Options Below)

Write a system of equations to solve the following problem Let c be the number of child tickets and a be the number of adult tickets Each child ticket for a rid class=
Write a system of equations to solve the following problem Let c be the number of child tickets and a be the number of adult tickets Each child ticket for a rid class=
Write a system of equations to solve the following problem Let c be the number of child tickets and a be the number of adult tickets Each child ticket for a rid class=
Write a system of equations to solve the following problem Let c be the number of child tickets and a be the number of adult tickets Each child ticket for a rid class=

Respuesta :

3C + 5A = 115
C + A = 33
A = 33 - C
3C + 5(33-C) = 115
3C + 165 - 5C = 115
-2C = -50
C=25
25+A=33
A=8

C=25, A=8



Answer:

3c + 5a=115

a+c=33

A=8

C=25

Step-by-step explanation:

In this case you just have to know that the price of the child tickets are $3, so thats the constant thas goes along with the C and the price for the adult tickets is $5 and that is the constant number that goes with the A, so the first equations would be:

3c + 5a=115

And you know that they sold 33 tickets so the total number of adult tickets added up to the total number of child tickets should be 33 so the second equation should be this:

a + c= 33

You just have to eliminate the c by multiplying the second equation by -3

3c + 5a=115  =       3c+ 5a= 115

(a + c= 33) -3=       -3c -3a= -99

You add both equations and it will end up like this:

+5a - 3a= 115-99

2a=16

A=8

So your first answer is A=8

and to solve the value for C you just put the value of A into one of the eqquations of the system:

A+C=33

C=33-8

C=25