Complicated algebra

  • Thread starter Brunno
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  • #1
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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]
 

Answers and Replies

  • #2
80
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What i meant was simplify the algebra to this one:

[tex]\frac{v_2[3v_1+v_2]}{[v_1+3v_2]}[/tex]
 
  • #3
1,233
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I don't think they are the same expression...
 
  • #4
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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]
 
  • #5
1,233
18
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.
 
  • #6
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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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