What's the rate of the stream's current? At his usual rate a man rows 15 miles downstream in five hours less than it takes him to return. If he doubles his usual rate, the time dowstream is only hour less than the time uqstream. What's the rate of the stream's current?
Let the man's rate in still water be $r$ and the rate of the current be $c$.
\begin{align*}
  \frac{15}{r-c}-\frac{15}{r+c}    &= 5 \tag{1}\\
  \frac{15}{2r-c}-\frac{15}{2r+c}    &= 1 \tag{2}
\end{align*}
Considering equation 1.
\begin{align*}
  \frac{15(r+c)}{(r-c)(r+c)}-\frac{15(r-c)}{(r+c)(r-c)}    &= 5 \\
  \frac{15r+15c}{(r-c)(r+c)}-\frac{15r-15c}{(r+c)(r-c)}    &= 5 \\
  \frac{15r+15c}{r^2-c^2}-\frac{15r-15c}{r^2-c^2}    &= 5 \\
  \frac{30c}{r^2-c^2}    &= 5 \\
   5r^2-5c^2     &= 30c \tag{3}
\end{align*}
Considering equation 2.
\begin{align*}
  \frac{15(2r+c)}{(2r-c)(2r+c)}-\frac{15(2r-c)}{(2r+c)(2r-c)}    &= 1 \\
  \frac{30r+15c}{(2r-c)(2r+c)}-\frac{30r-15c}{(2r+c)(2r-c)}    &= 1 \\
  \frac{30r+15c}{4r^2-c^2}-\frac{30r-15c}{4r^2-c^2}    &= 1 \\
  \frac{30c}{4r^2-c^2}    &= 1 \\
  4r^2-c^2     &= 30c \tag{4}
\end{align*}
We create a system of equations using 3 and 4.
\begin{align*}
   5r^2-5c^2     &= 30c \tag{3}\\
  4r^2-c^2     &= 30c \tag{4}
\end{align*}
\begin{align*}
   5r^2    &= 30c +5c^2\\
   \frac{5r^2}{5}    &= \frac{30c +5c^2}{5}\\
   \frac{5r^2}{5}    &= \frac{30c +5c^2}{5}\\
   r^2    &= 6c +c^2
\end{align*}
We subsitute into equation 4.
\begin{align*}
  4r^2-c^2     &= 30c \\
  4(6c +c^2)-c^2     &= 30c \\
  24c +3c^2    &= 30c \\
\frac{3c^2}{3c}    &= \frac{6c}{3c} \\
c &= 2
\end{align*}
I'm not feeling very confident of my answer.

This is correct. I would have made things a little simpler, noting that I have $5r^2-5c^2=30c=4r^2-c^2$, and so $r^2=4c^2$. Substituting $r=2c$ in either of these equations then gives the result a little more easily.
