Question 1031863
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frank went 16 miles at one speed and then came back going 4mph faster. if the return trip took 40 mins 
less time find the 2 speeds.
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Let x be the rate going to there, in miles per hour.

Then the rate going back is (x+4) miler per hour.


The time going at one speed is  {{{16/x}}}  hours.

The time going back is  {{{16/(x+4)}}}  hours.


The time equation is

    {{{16/x}}} - {{{16/(x+4)}}} = {{{2/3}}}.


To solve, multiply both sides by LCD  3x*(x+4).  You will get


    48(x+4) - 48x = 2x(x+4)     

    192 = 2x(x+4)

     96 = x(x+4)

    x^2 + 4x - 96 = 0

    (x+12)*(x-8) = 0


The roots are -12 and 8.  We reject negative root and accept the positive one x = 8.


<U>ANSWER</U>.  The speeds are 8 mph (going to there) and 12 mph (going back).


<U>CHECK</U>.  The time going to there is  {{{16/8}}} = 2 hours.

        The time going back is  {{{16/(8+4)}}} = {{{16/12}}} = {{{4/3}}} = 1{{{1/3}}} hours.

        The difference is  {{{2/3}}}  of an hour,  which is PRECISELY CORRECT.
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