[tex]\bf \begin{array}{ccccllll}
&distance&rate&time(hrs)\\
&-----&-----&-----\\
upstream&d&r-3&\frac{2}{3}
\\\\
downstream&d&r+3&1
\end{array}\\\\
-----------------------------\\\\
thus\qquad
\begin{cases}
d=(r-3)\frac{2}{3}
\\\\
d=(r+3)1
\end{cases}\qquad d=d\qquad thus
\\\\\\
(r-3)\cfrac{2}{3}=(r+3)1[/tex]
notice, the distance is the same, upstream or downstream, thus is "d" for both
solve for "r"
