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tickets \$7 in advance, \$9 at the door. 684 tickets were sold with a total \$5740. how many tickets were sold at the door?

In these types of problems, you will get two equations: the first equation relates to the dollars and the second equation relates to "how many".

OK, lets call advance tickets = a and door tickets = d. These are our unknowns.

The first equation is about the money. \$7a+\$9d=\$5740.

The second equation is about how many. a+d=684.

We have 2 equations and 2 unknowns, so we can use substitution method to solve this thing! Let's rewrite the second equation a=684-d (you could also solve this by writing d=684-a, you choose).

Substituting our new second equation into our first equation, we get

\$7(684-d)+\$9d=\$5740. Now using your skills to solve equations, here are the steps:

4788-7d+9d=5740

4788+2d=5740

2d=952

Sorry, WyzAnt just cut me off mid-solve!!

I was at 2d=952

d=476.

Therefore, 476 tickets were sold at the door.

9/24/2012

Mental Math approach:

Assuming all the tickets were sold in advance, then 684 tickets would have been sold with 7*684 = \$4788

Since the actual ticket revenue was \$5740, the difference 5740 - 4788 = \$952 would be the "extra" revenue.

Therefore, the actual number of tickets sold at the door was 952/2 = 476 tickets.

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