1) We can solve this problem, by considering that we are dealing with ratios. So let's start out writing the equation keeping in mind this relation:
[tex]rate=\frac{d}{t}[/tex]2) So, we can write out this keeping also in mind that Kerrie goes upriver and down the river:
[tex]\begin{gathered} Upriver:r-2mph \\ Downriver:\:r+2 \\ \frac{21}{r+2}+\frac{21}{r-2}=8 \\ \\ \frac{21}{r+2}\left(r+2\right)\left(r-2\right)+\frac{21}{r-2}\left(r+2\right)\left(r-2\right)=8\left(r+2\right)\left(r-2\right) \\ 21\left(r-2\right)+21\left(r+2\right)=8\left(r+2\right)\left(r-2\right) \\ 42r=8r^2-32 \\ 8r^2-32-42r=42r-42r \\ 8r^2-42r-32=0 \\ r_=\frac{-\left(-42\right)\pm\sqrt{\left(-42\right)^2-4\cdot\:8\left(-32\right)}}{2\cdot\:8} \\ r_1=5.925,r_2=-0.67 \\ \end{gathered}[/tex]We can discard negative values for the rate. Then the rate in still water is 5.9 mh
