Question 1132693
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<pre>
Let  "x"  be the average speed of the boat relative to the water (usually called as a "boat speed in still water"), in miles per hour.


Then the speed of the boat downstream is (x+4) mph, and the time traveling 48 miles downstream is  {{{48/(x+4)}}} hours.


     The speed of the boat   upstream is (x-4) mph, and the time traveling 48 miles   upstream is  {{{48/(x-4)}}} hours.


Total time for the round trip is 16 hours (given !), which gives you the "time" equation


    {{{48/(x-4)}}} + {{{48/(x+4)}}} = 16    hours.     (1)


It is your basic equation to solve.


At this point, I just know the answer: it is  8 miles per hour for the boat speed in still water.


But you, probably, want to get the solution on formal algebra way.  OK, let's do it.


Multiply equation  (1)  by  (x-4)*(x+4) = {{{x^2-16}}}. You will get


    48*(x+4) + 48*(x-4) = 16*(x^2-16).



Simplify it step by step.

    48x + 4*48 + 48x - 4*48 = {{{16x^2 - 256}}}

    {{{16x^2 - 96x - 256}}} = 0.


Factor out the common factor of 16 and cancel it.  You will get

    {{{x^2 - 6x - 16}}} = 0.


Factor left side

    (x-8)*(x+2) = 0.


The two roots of the last equation are  x= 8  and  x= -2.  In this problem only positive root  x= 8 is meaningful solution.


So, we got the same answer as I anticipated above.


<U>Answer</U>.  The boat speed in still water is 8 miles per hour.


<U>CHECK</U>.   Let's check equation (1). Its left side is

         {{{48/(8-4)}}} + {{{48/(x+4)}}} = {{{48/4}}} + {{{48/12}}} = 12 + 4 = 16 hours - same as the given total time.   ! The solution is correct !
</pre>

Solved.


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It is very standard and typical round trip upstream and downstream Travel and Distance problem.


See the lessons

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/travel/Wind-and-Current-problems.lesson>Wind and Current problems</A> 

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/travel/More-problems-on-upstream-and-downstream-round-trips.lesson>More problems on upstream and downstream round trips</A> 

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/word/travel/Wind-and-Current-problems-solvable-by-quadratic-equations.lesson>Wind and Current problems solvable by quadratic equations</A> 

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/travel/Unpowered-raft-moving-downstream-along-a-river.lesson>Unpowered raft floating downstream along a river</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/word/travel/Selected-problems-from-the-archive-on-a-boat-floating-Upstream-and-Downstream.lesson>Selected problems from the archive on the boat floating Upstream and Downstream</A> 

in this site, where you will find other similar solved problems with detailed explanations.


Read them attentively and learn how to solve this type of problems once and for all.


Also, &nbsp;you have this free of charge online textbook in ALGEBRA-I in this site

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-I - YOUR ONLINE TEXTBOOK</A>.


The referred lessons are the part of this textbook under the section "<U>Word problems</U>", &nbsp;the topic "<U>Travel and Distance problems</U>".



Save the link to this online textbook together with its description


Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson


to your archive and use it when it is needed.



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I read the &nbsp;"solution" &nbsp;by &nbsp;@stanbon and found it &nbsp;<U>totally wrong</U>.


For your safety, simply ignore it.


It is why I presented my solution here.