Single equation involving two functions

[Freye]

1.
2f(x) +3f(2010/x) = 5x
f(6) = ?



2.
No relevant equations that I am aware of (although I suspect that the answer requires some law or equation that I am don't know)



3.
2f(x) = 3f(2010/x) = 5x
2f(x) = 5x - 3f(2010/x)
f(6) = 5(6)/2 - 3f(2010/6)/2
f(6) = 15 - 3f(335)/2

I suspect that this isn't even close to what I am supposed to do, but I haven't seen a question like this before, and I don't know of any way to solve it completely without another equation so I can use substitution for f(335).

 

[Mark44]

1.
2f(x) +3f(2010/x) = 5x
f(6) = ?



2.
No relevant equations that I am aware of (although I suspect that the answer requires some law or equation that I am don't know)



3.
2f(x) = 3f(2010/x) = 5x
2f(x) = 5x - 3f(2010/x)
f(6) = 5(6)/2 - 3f(2010/6)/2
f(6) = 15 - 3f(335)/2

I suspect that this isn't even close to what I am supposed to do, but I haven't seen a question like this before, and I don't know of any way to solve it completely without another equation so I can use substitution for f(335).

Actually, I think you might be on the right track.
You have f(6) + (3/2) f(335) = 15

Now let x = 335 and plug that into your equation. That will give you two equations in the two unknowns f(6) and f(335), so you should be able to solve this system algebraically for f(6).

 

[zgozvrm]

Looks good to me...

 

[Mark44]

Looks good to me...

If you're referring to Freye's answer, f(6) is in terms of f(335), which is not known, so f(6) isn't known, either.

 

[Freye]

Thanks Mark, I actually thought of doing that at one point but for some reason I decided that I wasn't allowed to. I solved it and got the right answer! Thanks again