Question 133921
Rate of the boat in still water (the thing we want to determine): {{{r}}}


The time it takes for the boat to go upstream: {{{t[u]}}}


The time it takes for the boat to go downstream:  {{{t[d]}}}


The speed of the current: {{{5mph}}}


The distance traveled: {{{25m}}} each way.


When the boat is going upstream, it's rate relative to the shoreline is the rate of the boat in still water MINUS the rate of the current:  {{{r-5}}}.


Likewise, when the boat is going downstream, the rate relative to the shoreline is: {{{r+5}}}.


Using the relationship {{{d=rt}}}, we can write an equation for the upstream trip thusly:


{{{25=t[u](r-5)}}}


And we can write an equation for the downstream trip:


{{{25=t[d](r+5)}}}


However, we are given that the entire trip took 12 hours, so we know that {{{t[u]+t[d]=12}}}, which can be written {{{t[u]=12-t[d]}}}


Using this information we can re-write the equation for the upstream trip by substitution:


{{{25=(12-t[d])(r-5)}}}


Beginning with the equation for the downstream trip, solve for {{{t[d]}}}:


{{{t[d]=25/(r+5)}}}


And then solve the last form of the equation for the upstream trip for {{{t[d]}}}


{{{(12-t[d])=25/(r-5)}}}
{{{-t[d]=(25/(r-5))-12}}}
{{{t[d]=-(25/(r-5))+12}}}


Now we have two different expressions in terms of the rate of the boat in still water, {{{r}}}, that are equal to the time for the downstream trip {{{t[d]}}}, so set these two expressions to be equal to each other:


{{{25/(r+5)=-(25/(r-5))+12}}}


Simplify and isolate {{{r}}}
{{{25/(r+5)+(25/(r-5))-12=0}}}


Need a common denominator, which would be {{{(r+5)(r-5)}}}, so:
{{{(25(r-5)+25(r+5)-12(r+5)(r-5))/((r+5)(r-5))=0}}}


Distribute and collect like terms:
{{{(25r-125+25r+125-12(r^2-25))/((r+5)(r-5))=0}}}
{{{(25r-125+25r+125-12r^2+300)/((r+5)(r-5))=0}}}
{{{(-12r^2+50r+300)/((r+5)(r-5))=0}}}


Since we know that {{{a/b=0}}} if and only if {{{a=0}}} and {{{b<>0}}}, all we have to do is set the numerator equal to 0 and solve, excluding any root that would make the denominator go to zero, namely 5 and -5, and also excluding any root that is negative because a negative value for time would make no sense.


{{{-12r^2+50r+300=0}}}


We can take out a factor of -2 to make the coefficients a little more manageable:


{{{6r^2-25r-150=0}}}


Very conveniently, this quadradic is factorable:  (How did I know that?  I used an on-line quadratic solver -- there are a number of them available -- and noted that the roots were rational numbers.  I just worked backwards from the roots given by the solver to get the factors)


{{{(2r-15)(3r+10)=0}}}


Therefore:
{{{r=7.5}}} or {{{r=-10/3}}}


Neither root is 5 or -5, but the second one is negative and can be excluded.  The one acceptable solution is {{{r=7.5}}} mph.


Check the answer:
The boat went upstream for 25 miles at {{{7.5-5=2.5}}}mph, so the trip took {{{t[u]=25/2.5=10}}} hours.


The boat went downstream for 25 miles at {{{7.5+5=12.5}}}mph, so the return trip took {{{t[d]=25/12.5=2}}} hours.


{{{10 + 2=12}}} Answer checks.