SOLUTION: a theater has a seating capacity of 900 and charges $4 for children, $6 for students, and $8 for adults. at a certain screening with full attendance, there wre half as many adults

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: a theater has a seating capacity of 900 and charges $4 for children, $6 for students, and $8 for adults. at a certain screening with full attendance, there wre half as many adults       Log On


   



Question 339215: a theater has a seating capacity of 900 and charges $4 for children, $6 for students, and $8 for adults. at a certain screening with full attendance, there wre half as many adults as children and students combined. the receipts totaled $5600. how many children attended the show?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let number of children attending = c
Let number of students attending = s
Let number of adults attending = a
given:
Receipts from children = 4c
Receipts from students = 6s
Receipts from adults = 8a
(1) 4c+%2B+6s+%2B+8a+=+5600
There was full attendance, so
(2) c+%2B+s+%2B+a+=+900
(3) a+=+%281%2F2%29%2A%28c+%2B+s%29
--------------------------
There are 3 equations and 3 unknowns, so
it should be solvable
From (3):
(3) a+=+%281%2F2%29%2A%28c+%2B+s%29
c+%2B+s+=+2a
Now substitute into (2)
(2) 2a+%2B+a+=+900
3a+=+900
a+=+300
From (1):
(1) 4c+%2B+6s+%2B+8%2A300+=+5600
4c+%2B+6s+=+3200
From above,
c+%2B+s+=+600
6c+%2B+6s+=+3600
And by subtraction:
6c+%2B+6s+=+3600
-4c+-+6s+=+-3200
2c+=+400
c+=+200
and,
c+%2B+s+=+600
200+%2B+s+=+600
s+=+400
There were:
200 children
400 students
300 adults
check answer:
(1) 4c+%2B+6s+%2B+8a+=+5600
(1) 4%2A200+%2B+6%2A400+%2B+8%2A300+=+5600
(1) 800+%2B+2400+%2B+2400+=+5600
5600+=+5600
OK