A movie theater has a seating capacity of 235. 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 $ 1704, How many children, students, and adults attended?

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118 children

58 students

59 adults

Step-by-step explanation:

We can solve this problem by setting up a system of equations:

Let a = adults

2a = children (since double the # of adults were children), and

s = students

Set up the equations:

1704 = 5(2a) + 7s + 12(a)

1704 = 10a + 7s + 12a

235 = 2a + a + s

Simplify the equations:

1704 = 22a + 7s

235 = 3a + s

Subtract the bottom equation from the top by multiplying the bottom equation by 7 to eliminate the 's' variable:

1704 = 22a + 7s

7(235 = 3a + s)

1704 = 22a + 7s

1645 = 21a + 7s

(Subtract)

59 = a

This is the number of adults. Substitute this number into an equation to solve for the number of students:

235 = 3(59) + s

235 = 177 + s

s = 58.

Since the number of children is equivalent to 2a, solve:

2(59) = 118 children.

Therefore, the values for each group are:

118 children

59 adults

58 students.

adults: 59, students:58 and children 118

Step-by-step explanation:

let A for adults, and C = children and S for students

There are half as many adults as there are children=

A=C/2 , C=2A

A+C+S=235 or

A+2A+S=235 first equation

3A+S=235

12A+5C+7S =1704 or

12A+10A+7S=1704

22A + 7S=1704 second equation

3A+S=235 first

solve by addition and elimination

22A+7S=1704

21 A+7S=1645 subtract two equations

A=59 adults

C=2A=2(59)=118

substitute in :A+S+C=235

S=235-(118+59)=58

check: 5C+7S+12A=1704

5(118)+7(58)+12(59)=1704

50x5=$250

step-by-step explanation:

multiplication is multiplication.   it might feel harder depending on what level of math you are in, but generally it will be the same old stuff, while implementing some new concepts and methods of solving work.