Solve Complicated Algebra with Expert Help

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The discussion focuses on simplifying a complex algebraic expression involving variables v1 and v2. The original expression is compared to a simplified version, with participants debating whether they are equivalent. The simplification process is linked to a scenario involving three tourists and their travel speeds. Confusion arises regarding the accuracy of the simplifications and their relation to the initial problem. The conversation highlights the challenges of algebraic manipulation in the context of a practical problem.
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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.




\frac{v_2+3v_1^2-2v_1v_2}{3v_2^2-3v_1^2-2v_2v_1}
 
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What i meant was simplify the algebra to this one:

\frac{v_2[3v_1+v_2]}{[v_1+3v_2]}
 
I don't think they are the same expression...
 
But it came from this:



\frac{2(v_1+v_2)-(v_2-v_1)}{(v_2-v_1)}* \frac{v_2-v_1}{2(v_1+v_2)+(v_2-v_1)}
 
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.
 
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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