# Finding Two Unknown Values With Four Known Values

So, back in high school I learned the algorithm I needed to use in order to solve this problem. Unfortunately, I haven't needed the algorithm since high school and now I can't remember what it is!

I have a few known values, but two unknown values. I need to find the unknown. Here are the known values:

• Tickets Sold 176

• Adult Ticket $53 • Child Ticket$31

• Total Gross Revenue $6875.17 I need to find how many adults and how many children there were. I know I can't get an accurate answer because some sales had a 6% tax applied, but a best guess is better than nothing. Thanks in advance! - Are you sure about those values? Beause there's no way that summing up those children and adult tickets would leave you with cents in the total cash... Also, I don't think this is an algebraic geometry question. – Rodrigo Meireles Nov 9 '12 at 13:22 Is this algebraic geometry? – Amr Nov 9 '12 at 13:24 What is the tax? – peoplepower Nov 9 '12 at 13:30 @peoplepower, It's 6%, but that isn't going to help here as not all sales had that added on. – titanofold Nov 9 '12 at 13:34 @WillHunting, I assure it is a real question. This is the actual problem I'm faced with currently, and a problem I'm up against at least monthly. – titanofold Nov 9 '12 at 13:34 ## 2 Answers Assuming all the tickets had$6\%$tax applied, a child's ticket gives$\$32.86$ and an adult ticket gives $\$56.18$. If they were all children, you would get$\$5783.36$. Each child converted to an adult brings in an additional $\$23.32$, so we need to convert$\frac {6875.17-5783.36}{23.32} \approx 46.81$passengers to adults. If none of the tickets were taxed, we would need about$64.5$adults. So your range is$47$to$64$adults. - This is what I couldn't remember! Thank you very much for your answer. – titanofold Nov 9 '12 at 17:30 Number of passegers is$176$, number of childs is$c$number of adults is$a$. Then$176=a+c \\ 53a+31c=6875.17$Solve for$a$and$c$-Edit- Is the total price correct? - Yes, the total is correct. We have a rather loose system, which makes me not happy when I get problems like this. – titanofold Nov 9 '12 at 13:30 As to your answer, I have to iterate through the possible solutions, or is there another method? I can't help but think there's another method. – titanofold Nov 9 '12 at 13:32 the problem is that$a$and$c\$ are natural numbers, and so are the prices. If you multiply natural numbers you get a natural number, and if you add natural numbers you get a natural number. But the price is not a natural number, so there must be something wrong somewhere. – Jorge Nov 9 '12 at 13:34
you are right. Unfortunately, I can't fix the thing that's wrong: the system in use. – titanofold Nov 9 '12 at 14:12