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

Question

I need help with this problem

Answer 1

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

 

Answer 2

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.

Answer 3

684=x+y

$5740=7x+9y

684(7)=7x+7y

5740-4788=2y

y= 952/2= 476 sold at the door

x= 684-476= 208 sold in advance