let's say "p" people were going to the expedition initially, and the cost for each was "c", now, we know the total cost is 1800, so for "p", folks that'd be 1800/p how much each one cost, namely, how many times "p" goes into 1800.
well, prior to leaving, 15 dropped out, so that leaves us with " p - 15 ", and the cost "c" bumped up to " c + 27 " for each.
[tex]\bf \begin{cases}
\cfrac{1800}{p}=\boxed{c}\\\\
\cfrac{1800}{p-15}=c+27\\\\
----------\\
\cfrac{1800}{p-15}=\boxed{\cfrac{1800}{p}}+27
\end{cases}\\\\
-------------------------------\\\\
\cfrac{1800}{p-15}={\cfrac{1800}{p}}+27\impliedby
\begin{array}{llll}
\textit{let's multiply both sides by}\quad p(p-15)\\
\textit{to get rid of the denominators}
\end{array}[/tex]
[tex]\bf 1800p=1800(p-15)+27[p(p-15)]
\\\\\\
1800p=1800p-27000+27(p^2-15p)
\\\\\\
0=-27000+27(p^2-15p)\implies 0=-27000+27p^2-405p
\\\\\\
\textit{now, let's take a common factor of }27
\\\\\\
0=p^2-15p-1000\implies 0=(p-40)(p+25)\implies p=
\begin{cases}
\boxed{40}\\
-25
\end{cases}[/tex]
well, you can't have a negative value of people... so it has to be 40.
so, 40 folks were initially going, then 15 dropped out, how many went on the expedition? 40 - 15.
