Question 172638
I set up the equations you solve
:
d=rt........in both instances distance is equal
lets call the rate of the slow trip r and time t
lets call the rate of the fast trip r+10 and time(t-1)
:
so 200=rt...eq 1
...200=(r+10)(t-1)..eq 2
:change eq 1 to t=200/r and substitute it into eq 2 and solve for r...this will involves a quadratic equation again
:
{{{200=(r+10)((200/r)-1)}}}
:
{{{200=(r+10)((200-r)/r)}}}
:
{{{200r=(r+10)(200-r)}}}
:
{{{200r=200r-r^2+2000-10r}}}
:
{{{r^2+10r-2000=0}}}

throw out negative value{{{system(r=40,r=-50)}}}
so {{{highlight(r=40)}}}mph and {{{highlight(r+10)}}}=50mph
:

 *[invoke quadratic "r", 1, 10, -2000]

:
:
:
:
Let s= speed of the boat in still water
The current =3 mi/hr
we know that d=rt or d/r=t for this problem we must break this up into 2 parts
(distance upriver)/(rate going upriver)
+ (distance downriver)/(rate going downriver) = 9 hrs
:the we know the distance is 60 each way
the rate going up stream is 60/s-3
the rate going down stream is 60/s+3
:
so we have {{{60/(s-3)+60/(s+3)=9}}}again a quadratic equation
:
:
multiply each term by (s+3)(s-3)
:
{{{60(s+3)+60(s-3)=9(s+3)(s-3)}}}
{{{60s+180+60s-180=9(s^2-9)}}}
{{{120s=9s^2-81}}}
{{{9s^2-120s-81=0}}}
:
drop the negative value{{{system(s=13.98,s=-.64)}}}
:
so {{{highlight(s=14)}}} mph


*[invoke quadratic "s", 9, -120, -81]