A crew can now row a certain course up the stream in $84$ minutes; they can row the same course down stream in $9$ minutes less than they can row it in still water. How long would they take to row down with the stream?
My solution approach :-
Let the speed of the boat be $x$ units.
Let the speed of the stream be $y$ units.
Let the distance be $d$ units.
Upstream time = $\frac{d}{x-y} = 84$ minutes
Downstream time = $\frac{d}{x+y}$
Still water time = $\frac{d}{x}$
As per the question :-
$\frac{d}{x}$ - $\frac{d}{x+y}$ = $9$
Now I am getting stuck after this as this now has $3$ variables and only $2$ equations and how can we find the value of all the variables. I do not see any way ahead. Is the question incomplete? Please help me with this!!!
Thanks in advance !!!
Notice how all the terms have $d$ in the numerator. This motivates us to treat $d$ as a constant and solve for $x,y$ in terms of $d$. Dividing by $d$ on both sides, we have $$\begin{cases} \frac{1}{x-y}=\frac{84}{d}\\ \frac{1}{x}-\frac{1}{x+y}=\frac{9}{d}\end{cases}$$ $$\begin{cases} \frac{1}{x-y}=\frac{84}{d}\\ \frac{y}{x(x+y)}=\frac{9}{d}\end{cases}$$ Since we only want to find $\frac{d}{x+y}$. Make the substitution $u=x+y$, and then put everything in terms of $u$ and $y$. $$\begin{cases} \frac{1}{u-2y}=\frac{84}{d}\\ \frac{y}{u(u-y)}=\frac{9}{d}\end{cases}$$ Dividing the two equations gives $$\frac{u(u-y)}{y(u-2y)}=\frac{28}{3}$$ $$3u^2-3uy=28uy-56y^2$$ $$3u^2-31uy+56y^2=0$$ $$(3u-7y)(u-8y)=0$$ $$u=\frac{7}{3}y,8y$$ Since the speed of the boat should be faster than the stream (so that they can travel upstream), we have $$x>y$$ $$x+y>2y$$ $$u>2y$$ Hence, both of these answers are possible
Case 1: $u=\frac{7}{3}y$ Substituting back into our first equation, we have $$\frac{1}{u-2y}=\frac{84}{d}$$ $$d=84(u-2y)$$ $$d=84(\frac{1}{3}y)$$ $$d=28y$$ Hence, $$\frac{d}{u}=\frac{28y}{\frac{7}{3}y}=\boxed{12}$$
Case 2: $u=8y$ Substituting back into our first equation, we have $$\frac{1}{u-2y}=\frac{84}{d}$$ $$d=84(u-2y)$$ $$d=84(6y)$$ $$d=504y$$ Hence, $$\frac{d}{u}=\frac{504y}{8y}=\boxed{63}$$
I guess both of these answers are acceptable then, as I can't find any reason to conclude either as extraneous.