Question 199535
Let x = speed of boat in still water
 
 
 
{{{d=rt}}} Start with the distance-rate-time formula
 
 
{{{40=(x-3)t}}} Plug in {{{d=40}}} and {{{r=x-5}}}. This equation represents the upstream journey.
 
 

{{{40/(x-5)=t}}} Divide both sides by {{{x-5}}} to isolate "t";
 


So the expression for the time it takes to go upstream can be represented by the expression {{{40/(x-5)}}}


 
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{{{d=rt}}} Go back to the distance-rate-time formula

 
{{{40=(x+5)t}}} Plug in {{{d=40}}} and {{{r=x+5}}}. This equation represents the downstream journey
 

 
{{{40/(x+5)=t}}} Divide both sides by {{{x+5}}} to isolate "t";
 

 
So the expression for the time it takes to go downstream can be represented by the expression {{{40/(x+5)}}}
 
 

Now simply add the two time expressions to get: {{{40/(x-5)+40/(x+5)}}}
 


{{{40/(x-5)+40/(x+5)=6}}} Now set that expression equal to the total time of 6 hours
 

 
{{{40(x+5)+40(x-5)=6(x+5)(x-5)}}} Multiply <b>every</b> term by the LCD {{{(x+5)(x-5)}}} to clear the denominators.
 

 
{{{40(x+5)+40(x-5)=6(x^2-25)}}} FOIL
 

 
{{{40x+200+40x-200=6x^2-150}}} Distribute
 

 
{{{40x+200+40x-200-6x^2+150=0}}} Subtract {{{6x^2}}} from both sides. Add {{{150}}} to both sides.
 
 

{{{-6x^2+80x+150=0}}} Combine like terms



Notice we have a quadratic in the form of {{{Ax^2+Bx+C}}} where {{{A=-6}}}, {{{B=80}}}, and {{{C=150}}}



Let's use the quadratic formula to solve for "x":



{{{x = (-B +- sqrt( B^2-4AC ))/(2A)}}} Start with the quadratic formula



{{{x = (-(80) +- sqrt( (80)^2-4(-6)(150) ))/(2(-6))}}} Plug in  {{{A=-6}}}, {{{B=80}}}, and {{{C=150}}}



{{{x = (-80 +- sqrt( 6400-4(-6)(150) ))/(2(-6))}}} Square {{{80}}} to get {{{6400}}}. 



{{{x = (-80 +- sqrt( 6400--3600 ))/(2(-6))}}} Multiply {{{4(-6)(150)}}} to get {{{-3600}}}



{{{x = (-80 +- sqrt( 6400+3600 ))/(2(-6))}}} Rewrite {{{sqrt(6400--3600)}}} as {{{sqrt(6400+3600)}}}



{{{x = (-80 +- sqrt( 10000 ))/(2(-6))}}} Add {{{6400}}} to {{{3600}}} to get {{{10000}}}



{{{x = (-80 +- sqrt( 10000 ))/(-12)}}} Multiply {{{2}}} and {{{-6}}} to get {{{-12}}}. 



{{{x = (-80 +- 100)/(-12)}}} Take the square root of {{{10000}}} to get {{{100}}}. 



{{{x = (-80 + 100)/(-12)}}} or {{{x = (-80 - 100)/(-12)}}} Break up the expression. 



{{{x = (20)/(-12)}}} or {{{x =  (-180)/(-12)}}} Combine like terms. 



{{{x = -5/3}}} or {{{x = 15}}} Simplify. 



So the <i>possible</i> solutions are {{{x = -5/3}}} or {{{x = 15}}} 

  

However, you can't have a negative speed. So the only answer is {{{x = 15}}}



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Answer:



So the speed of the boat in still water is 15 mph.