Question 187282
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Felipe jobs for 10 miles and then walks another 10 miles.  He jobs 2 1/2 miles per hour faster than he walks, and
the entire distance of 20 miles takes 6 hours.  find the rate at which he walks and the rate at which he jogs.
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The other person's solution is rather lengthy, and in this author's opinion, a few UNNECESSARY variables were introduced. 
Furthermore, he suggests using the quadratic formula to solve his final equation, which can be easily FACTORIZED. 

Let the speed at which he walks, be S
Then, his jogging speed = {{{S + 2&1/2}}}, or S + 2.5
Time taken to walk 10 miles: {{{10/S}}}, and time taken to jog 10 miles = {{{10/(S + 2.5)}}}
As he takes 6 hours to walk and jog 20 miles, we get the following
TOTAL-TIME equation: {{{10/S + 10/(S + 2.5) = 6}}}
                              10(S + 2.5) + 10S = 6S(S + 2.5) ----- Multiplying by LCD, S(S + 2.5)
                                     {{{10S + 25 + 10S = 6S^2 + 15S}}}
                                              {{{20S + 25 = 6S^2 + 15S}}}
                           {{{6S^2 + 15S - 20S - 25 = 0}}}
                                      {{{6S^2 - 5S - 25 = 0}}}
                          {{{6S^2 - 15S + 10S - 25 = 0}}}
                        {{{3S(2S - 5) + 5(2S - 5) = 0}}}
                               (2S - 5)(3S + 5) = 0
                                                 2S - 5 = 0      or     3S + 5 = 0 ---- Setting FACTORS equal to 0
                                                      2S = 5      or            3S = - 5 (IGNORE)
                  <font color = red><font size = 3>Walking speed</font></font>, or S = {{{5/2}}} = <font color = red><font size = 3>2.5 mph</font></font>

                 <font color = red><font size = 3>Jogging speed: </font></font>S + 2.5 = 2.5 + 2.5 = <font color = red><font size = 3>5 mph</font></font>

                 You can do the CHECK!!</font></font></font></b></pre>