The school is having a play at the theatre it cost 1 cent for children, 5 cents for teenagers, 10 cents for adults , there was 100 people that showed up and only 2 dollars and 80 cents earned, how many children teens and adults showed up for the school play?
There are multiple solutions to this Diophantine equation:
55 children, 45 teenagers, 0 adults
60 children, 36 teenagers, 4 adults
65 children, 27 teenagers, 8 adults
70 children, 18 teenagers, 12 adults
75 children, 9 teenagers, 16 adults
80 children, 0 teenagers, 20 adults
Is this a homework question? (Asking for help is OK. Asking to have it done is not.) That said, I will continue because there is not unique solution. This is because there are two equations and three unknowns. Remember how to set it up into equation form? You should get this:
(0.01c) + (0.05t) + (0.10*a) = 2.80
c + t + a = 100
Pick a number for either c, t or a and then you can practice solving it. (This would be the third equation, such as c = 80.) Use some of the numbers that Giles provided.
But, remember, if you cannot generate the same number of equations and unknowns, you cannot get a unique answer.
This is a set of Diophantine equations, which requie that the answers all be non-negative integers. With such equations, it is possible to have a unique solution for n unknowns with less than n equations.
Example: The concert charges 2 cents for children and 3 cents for adults. The total takings were 5 cents. How many children and how many adults? The equation
2x + 3y = 5
has a (trivial) unique solution if x and y must both be non-negative integers.
thank you, Mycroft, that was exactly the the solution for that problem, The problem was I did not check to see if there was multiple solutions to the answer, If there had not been it would have taken you a lot longer. I will try to figure out the original numbers for it so there will only be 1 answer.
Because I think it was last year sometime it was homework and it literally took me forever.
Giles, I stand corrected. It’s been way too long since studying maths and if you can reel off a phrase such as “Diophantine equation,” I will definitely bow to your superior knowledge. 
(My knowledge level is very basic. Plus I forgot about the entire integer thing. We don’t want any fractional people.)
JustWrightin, a beginning (learning) way to solve something like this is to pick varying values for one of the unknowns. In this case use ‘a’ because you will have less choices. Start with a = 0, then a=1, a=2, … up to a=28. (Extra credit if you determine why to stop there.) Then solve for the other two equations. It will take a while, but not forever, because you will only look at whole numbers for answers. As I said before, we don’t want fractional people.