A man jumped into his canoe and paddled upstream for one mile. After this, he continued for another fifteen minutes. Having arrived at his destination, he then turned around and paddled downstream, arriving at his starting point in exactly one hour. Assuming that it flowed at a constant rate, how fast was the river current?
my attempt:
m = mans speed,
r = river speed,
t = time coming back,
then
distance going = distance coming, thus
$1+(m-r)(15/60) = (m+r)t$ also for the first mile the distance 1 was traveled in 1 - t - 15/60 hours thus.. $$1=(m-r)(1-t-15/60)$$
combining these to turn $$1+(m-r)(15/60) = (m+r)t$$ into
$$(m-r)(1-t-15/60)+(m-r)(15/60) = (m+r)t$$
or
$$(m-r)(1-t) = (m+r)t$$
This leaves too many degrees of freedom for a unique solution.
I am not sure if I've missed something. Any help would be appreciated.
edit: Second Attempt
m = speed of man
r = speed of river
t = time for first mile (slightly different)
FirstMile = $(m-r)*t$;
Next15Minutes = $(m-r)*15/60$;
TheReturn = $(m+r)*(1-15/60-t)$;
$$m t - r t + \frac{1}{4} \, m - \frac{1}{4} \, r = -m t - r t + \frac{3}{4} \, m + \frac{3}{4} \, r$$
$$4 \, m t - 4 \, r t + m - r = -4 \, m t - 4 \, r t + 3 \, m + 3 \, r$$
$$0 = -8 \, m t + 2 \, m + 4 \, r$$
$$r=\frac{1}{2} \, m {\left(4 \, t - 1\right)}$$
was hoping to argue that $t\ge 1/4$ and $t\le 1/4$ that would resolve it.. but only $t\ge 1/4$ is clear.
Your setup looks fine to me, though perhaps you have not used your equations to best advantage - at the end you have derived another equation but you haven't eliminated a variable, which would probably be the best thing to do.
I can't see that there are any conditions on the problem that you have missed.
So we can check that there is no unique solution by giving two solutions. By trial and error, $$t=\tfrac14\ ,\quad m=4\ ,\quad r=2\ ,$$ or $$t=\tfrac13\ ,\quad m=\tfrac{18}5\ ,\quad r=\tfrac65\ .$$
This seems pretty unusual for a school problem, I would check with the student to make sure that no conditions have been missed.
Note that even if the system as a whole does not have a unique solution, there may be a unique solution for one of the variables. For example, $$r+t+m=5\ ,\quad r-t-m=3$$ gives $2r=8$ and hence $r=4$, though there is no unique solution for $m$ and $t$. However we see above that your problem is not like this.