frog question

Hey ant101!

x = speed of frog

y = speed of current

We use (distance)=(rate)*(time) 

We can set up the system of equations:

\((x+y)\cdot2=8\\ (x-y)\cdot14=8\)

Solving for x, we get:

\(x=\frac{16}7\)

I hope this helped,

Gavin

The current helps the frog swim downstream and hinders her swimming upstream. Let the frog's rate in still water be \(x\) and the current be \(y.\) Thus, the swimming downstream the frog's rate is \(x+y\) , while swimming upstream it is \(x-y\) . Now apply rate times equals distance: downstream, we have 

                    \((x+y)(2)=8\)

and upstream gives

                    \((x-y)(14)=8\)

Solving these equations, we find that \((x,y)=(16/7, 12/7).\)

Thus the frog's rate in still water is \(\boxed{\frac{16}{7}}\)miles per hour.

Thanks so much!