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!