recall your d = rt, distance = rate * time.
so, let's say hmmm not using xx, let's say his speed going downstream is "r".
Now, notice, going downstream and coming back, is the same distance, say "d" miles.
the travel downstream took 3 hours the one upstream took 4 hours.
[tex]\bf \begin{array}{lccclll}
&\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\
&------&------&------\\
Downstream&d&r&3\\
Upstream&d&r-1&4
\end{array}
\\\\\\
\begin{cases}
d=3r\implies \frac{d}{3}=r\\
d=4(r-1)\\
--------\\
d=4\left(\boxed{\frac{d}{3}}-1 \right)
\end{cases}
\\\\\\
d=\cfrac{4d}{3}-4\implies d=\cfrac{4d-12}{3}\implies 3d=4d-12\implies 12=4d-3d
\\\\\\
12=d[/tex]
now, the kayaker went forth and back, so, the distance is d + d.
