Need help on an Algebra problem!! :DD

There are three teams that are working on a project.

Teams A&B together:
-6 days to complte
-Spent $8700

Teams B&C together:
-10 Days to complete
-Spent $9500

Teams A&C together:
-5 days to complete TWO-THIRDS
-spent $5500


Questions!
1. How much time does each individual team take on building the project by themselves.
2. If it was required to complete the project in less than 15 days, which team would spend the least amount of money?

Please write an algebraic equation to solve for this, and explain. Thank you sooo much! This is really confusing me for some reason T_T;

To find the time, here is your basic equation: (Group 1 + Group 2) = t, where t = time. S:
A+B = 6
B+C = 10
A+C = 5

Take first equation and find A and you get: A = 6-B
For second equation, find C and you get: C = 10-B
Plug them into third equation and you get: (6-B)+(10-B) = 5
Solve for B and you get 16-2B = 5, simplify and you you get B = 5.5 days. Now that you know B, you can find out the time for A and C, which is .5 and 4.5, respectively. I'm not clear on question 1, but I assume that each team does 1/2 of the project? So you just double the time. So if B were to do the project themselves, it would take them 11 days. A would take 1 day, and C would take 9 days.

Your basic equation to find money is t(Group1+Group2) = m, where t = time and m = money spent. So:
6(A+B) = 8700
10(B+C) = 9500
5(A+C) = 5500

So take your first equation and find A to get: A = 1450-B.
The second eq. and find C to get: C = 950-B
Plug into third eq. to get: 5(1450-B+950-B) = 5500
Solve and you get B = 650. So B uses $650
From here, I think you can find out how much A and C uses. And again, I think you'll have to double the amount since we want to know how much they -would spend- if the teams worked individually. Then to answer your 2nd question, just find which team uses the least amount of money.

I hope that helps. ^_^ And I know what you mean by confusing. It's not very clear problem that you have to make assumptions about it...and it's not good to assume in math. ^^;;; So I hope I did it right myself.... o_O

For the first problem, think of the teams as variables.
Teams A&B
6 days
6 days=144 hours
A+B=144

Teams B&C
10 days
10 days =240 hours
B+C=240

Teams A&C
5 days for 2/3
5 days=120 hours
120 divided by 2 is 60
60 hours for 1/3 so 180 hours for 3/3
A+C=180

Now you rearange them and plug them in.
Let's figure out A first
A+B=144
B+C=240-----B=240-C
A+C=180-----C=180-A

A+B+B+C+A+C=564
A+240-C+240-C+C+A+C=564
A+240-(180-A)+240-(180-A)+180-A+A+180-...
A+240-180+A+240-180+A+180-A+A+180-A=56... the same as x-1+y
2A+480=564
2A=84
A=42

A+C=180
42+C=180
C=138

B+C=240
B+138=240
B=102

Team A takes 42 hours to complete
Team B takes 102 hours to complete
Team C takes 138 hours to complete