Three adults and four children must pay $ 79. Two adults and three children must pay $ 56.
(1) A standard algebraic solution....
3a+4c = 79
2a+3c = 56
Multiply both equations by appropriate numbers so the number of children is the same in both (or you could make the numbers of adults the same). Then subtract the two equations to find the cost of each adult ticket.
9a+12c = 237
8a+12c = 224
------------
a = 13
The cost of each adult ticket is $13.
Substitute a=13 in either original equation to find c, the cost of each children's ticket.
2(13)+3c = 56
26+3c = 56
3c = 30
c = 10
The cost of each children's ticket is $10.
(2) An informal solution, using logical reasoning....
(a) The difference between the two purchases is 1 adult and 1 children's ticket for a total of $23.
(b) Double that: 2 adults and 2 children cost $46.
(c) The given information is that 2 adults and 3 children cost $56.
(d) Compare (b) and (c) to find the cost of one children's ticket is $10.
(e) Compare (a) and (d) to find the cost of one adult ticket is $13.
(3) A non-standard algebraic solution, using the path shown in the informal solution (2)....
Find the difference between the two original equations:
3a+4c = 79 [1]
2a+3c = 56 [2]
----------
a+ c = 23 [3]
Double that:
2a+2c = 46 [4]
Compare [2] and [4]
2a+3c = 56
2a+2c = 46
----------
c = 10 [5]
Use [3] and [5] to find a
a + c = 23
c = 10
----------
a = 13
