## Precalculus (6th Edition) Blitzer

$3$ miles/hour, $1.5$ miles/hour.
Step 1. Assume the average rowing velocity in still water is $x$ miles/hour and the average velocity of the current is $y$ miles/hour. Step 2. When rowing with the current, we have a total velocity of $x+y$ and $2(x+y)=9$ or $x+y=\frac{9}{2}$ miles. Step 3. When rowing against the current, we have a total velocity of $x-y$ and $6(x-y)=9$ or $x-y=\frac{3}{2}$ miles. Step 4. Adding the two equations above, we have $2x=6$ and $x=3$ miles/hour; thus $y=\frac{9}{2}-3=\frac{3}{2}=1.5$ miles/hour.