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# MET charges \$21 adults, \$14 seniors, and \$6 children. Revenue was \$9486 for 718 tickets. There were 44 more children than adults. How many children attended?

a) 258

b) 214

c) 246

d) 248

number of children = x+44

number of seniors = total number of tickets - adults - children = 718-x-(x+44)=674-2x

Ticket sales = \$21*adults + \$6*children +\$14*seniors

9486 = 21x + 6(x+44) + 14(674-2x)

9486 = 21x + 6x + 264 + 9736 - 28x

9486 = -x + 9700

214 = x = number of adults

number of children = x+44 = 214+44 = 258

3/5/2013

In this problem, you have three unknowns ( number of adults = A, number of children = C, and number of seniors = S).  To solve an equation with three unknowns, you need three equations minimum.  The first two equations come from the revenue and the number of tickets sold.  If the total number of tickets sold was 718, then the sum of the tickets sold for adults, children, and seniors must equal 718

A + C + S = 718

For the revenue, you need to use the price of each ticket multiplied buy the number of people in that price range:

21*A + 14*S + 6*C = 9486

For the last equation, use the last piece of information given: there are 44 more adults than children:

A = 44 + C

To reduce this three equation problem to two variables and two equations, substitute the last equation into the first two:

(44+C) + C + S = 718

21*(44+C) + 14*S + 6*C = 9486

After simplifying the expressions, you now have a system of equations like you've seen before.

number of children = x+44

number of seniors = total number of tickets - adults - children = 718-x-(x+44)=674-2x

Ticket sales = \$21*adults + \$6*children +\$14*seniors

9486 = 21x + 6(x+44) + 14(674-2x)

9486 = 21x + 6x + 264 + 9736 - 28x

9486 = -x + 9700

214 = x = number of adults

number of children = x+44 = 214+44 = 258