i got a tricky one for you guys. bobs friends decided to equally share $270 cost of a cookware set for a wedding gift.however before the wedding 3 more of bobs friends joined the others in giving the couple the cookware. this lowered the contibutuin of each by $15. how many friends ORIGINALLY agreeed to share the cost of it.

Question

accessdenied · Answer

So, I started with two definitions.
 $ n := \text{the number of original friends} $,
 $ C := \text{original amount paid by each friend.} $

We can then create two equations relating $n$ and $C$.
   $ \Large \frac{270}{n} = C $
 > Each friend is paying an equal amount, so it is evenly distributed to each friend.
   $ \Large \frac{270}{n+3} = C - 15 $
 > When three more friends join in, the amount paid by each goes down by 15.

Then, I substituted the first equation into the second to create a proportion. 
   $ \Large \frac{270}{n+3} = \frac{270}{n} - 15 $
   $ \Large \frac{270}{n+3} = \frac{270}{n} - \frac{15n}{n} $
   $ \Large \frac{270}{n+3} = \frac{270 - 15n}{n} $

I multiply by the lowest common denominator between the values, n(n+3) on both sides (or, essentially cross multiplication)
   $ \large 270n = (270 - 15n)(n + 3) $
   $ \large 270n = 270n + 810 - 15n^2 - 45n $
   $ \large 0 = -15n^2 - 45n + 810 $

We can factor out a -15 and divide it off.
   $ \large -15(n^2 + 3n - 45) = 0 $\
   $ \large n^2 + 3n - 45 = 0 $
 
Then factor and solve for $n > 0$ and $n$ are not restricted in the original equations:
   $ \large (n + 9)(n - 6) = 0 $
   $ \large n + 9 = 0 \text{ and } n - 6 = 0 $
   $ \large \cancel{n = -9} \text{ and } \boxed{n = 6} $

Thus, the original number of friends was $ 6 \text{ friends} $.

accessdenied · Answer

Of course, as I check, it does appear to work out as well.  Letting $n=6$, 
$$ \frac{270}{n} = \frac{270}{6} = 45 $$
$$ \frac{270}{n+3} = \frac{270}{9} = 30 $$ which is 15 less than the original.

anonymous · Answer

$$\frac{270}{n}-15\text{ = }\frac{270}{n+3}$$n = 6 or -9, n=6