Remember that [tex]Speed= \frac{distance}{time} [/tex]. Since the distance of the complete trip is 312 miles, the distance of each way is [tex] \frac{312}{2} =156[/tex] miles.
Let [tex]S _{b} [/tex] be the speed of the boat in still water and [tex]S_{c} [/tex] the speed of the current.
Now, for the downstream trip the boat is traveling with the current, so:
[tex]S_{b} +S_{c} = \frac{156}{13} [/tex]
[tex]S_{b} +S_{c} =12[/tex] this will be our equation (1)
For the trip back the boat is traveling against the current, so:
[tex]S _{b} -S_{c} = \frac{156}{26} [/tex]
[tex]S_{b} -S_{c} =6[/tex] this will be our equation (2)
Next, lets add equation (1) and equation (2) to get rid of [tex]S _{c} [/tex]:
[tex] \left \{ {{S_{b}+S_{c} =12} \atop {+(S_{b}-S_{c} =6})} \right. [/tex]
[tex]2S_{b} =18[/tex]
[tex]S _{b} = \frac{18}{2} [/tex]
[tex]S _{b} =9[/tex]
Finally, now that we know the speed of the boat in still water, lets replace that value in our equation (1) to find the speed of the current:
[tex]9+S_{c} =12[/tex]
[tex]S_{c} =12-9[/tex]
[tex]S_{c} =3[/tex]
We can conclude that the speed of the boath in still water is 9[tex] \frac{mi}{h} [/tex], and the current of the stream is 3[tex] \frac{mi}{h} [/tex].
