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Complicated algebra

  1. Mar 16, 2010 #1
    There's somebody who could help me solve this problem.It is from a very long resolution that i was trying to understand and this part i didn't get.




    [tex]\frac{v_2+3v_1^2-2v_1v_2}{3v_2^2-3v_1^2-2v_2v_1}[/tex]
     
  2. jcsd
  3. Mar 16, 2010 #2
    What i meant was simplify the algebra to this one:

    [tex]\frac{v_2[3v_1+v_2]}{[v_1+3v_2]}[/tex]
     
  4. Mar 16, 2010 #3
    I don't think they are the same expression...
     
  5. Mar 16, 2010 #4
    But it came from this:



    [tex]\frac{2(v_1+v_2)-(v_2-v_1)}{(v_2-v_1)}[/tex]* [tex]\frac{v_2-v_1}{2(v_1+v_2)+(v_2-v_1)}[/tex]
     
  6. Mar 16, 2010 #5
    That final expression does not equal either of the first two I don't think, though it simplifies to something similar to the second one (numerator is 3*v1 + v2), not v2*(3*v1 + v2).

    Where are you quoting from? None of this follows to me.
     
  7. Mar 16, 2010 #6
    It is from a old question:

    Three tourists gathered in one place and having a bike that can carry only two people ever need to get to a tourist destination as quickly as possible.The A tourist A takes tourist B, cycle to a point x of the course and returns to tourist C as he was walking to find A. tourist B from x continues to walk his journey to the tourist center.
    The three arrive simultaneously to the turistic centre.
    A average speed v1 is as pedestrian and cyclist as v2 average speed that the tourists will the total route.
     
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