SOLUTION: Please help me with this problem and show all work. The tutor that did the problem confused me even more. A movie theater has a seating capacity of 333. The theater charges $5.0

Question 1001677: Please help me with this problem and show all work. The tutor that did the problem confused me even more.
A movie theater has a seating capacity of 333. The theater charges $5.00 for children, $7.00 for students, and $12.00 of adults. There are half as many adults as there are children. If the total ticket sales was $ 2422, How many children, students, and adults attended?

Answer by macston(5194) (Show Source): You can put this solution on YOUR website!
.
C=number of children; S=number of students; A=number of adults
C=2A (half as many A's as C's)
.
C+S+A=333 . (Number of tickets=number of seats)
2A+S+A=333 . (This substitutes C=2A from above)
3A+S=333
S=333-3A (Use this to sub for S below)
.
$5C+$7S+$12A=$2422$5C+$7S+$12A=$2422(This converts number of tickets to dollars)
$5(2A)+$7S+$12A=$2422 (Substitute C=2A again)
$10A+$7(333-3A)+$12A=$2422 (Sub for S from above)
$10A+$2331-$21A+$12A=$2422
A=111
ANSWER 1: There were 111 adults.
.
S=333-3A=333-3(111)=333-333=0
ANSWER 2: There were no students.
.
C=2A=2(111)=222
ANSWER 3: There were 222 children.
.
CHECK:
C+S+A=333
222+0+111=333
333=333
.
$5C+$7S+$12A=$2422
$5(222)+$7(0)+$12(111)=$2422
$1110+$0+$1332=$2422
$2442=$2442
Hope this helps . Ask if you need more clarification.

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