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Hi, According to wolfram alpha x+y/ x-y can likewise be stood for as <-(2x)/(y-x)> -1 When it remains in this type it benefits me I simply can"t appear to obtain it because type, any kind of aid (or tips or standard policies to use) would certainly be substantially valued.
Sorry I simply understood this remains in the incorrect area, it is for a limitation concern however the inquiry itself is not. (I put on"t recognize just how to relocate this right into the ideal area)
We intend to share the specific (x+y)/(x-y) as <-2 x/ (y-x)> - 1. In order to create it right into that kind we will certainly initially secure -1 typical from the common denominator to obtain(x+y)/(x-y) = - (x+y)/ (y-x). Currently separate the numerator by the , ratio being 1 as well as rest 2x. Therefore we obtain: (x+y)/(x-y) = - (x+y)/ (y-x) = - <1 + 2x/ (y-x)> = -1 - 2x/ (y-x)
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Responses:1 individual
Thanks for the fast reply, however I have no concept exactly how you receive from:"Currently split the numerator by the common denominator, ratio being 1 as well as rest 2x. Therefore we obtain: - (x+y)/ (y-x) = - <1 + 2x/ (y-x)> = -1 - 2x/ (y-x)"I sanctuary"t done algebra in 4-5 years, could you please describe this detailed? (as I have actually failed to remember "standard mathematics policies")It"s - (x+y)/ (y-x) = - <1 + 2x/ (y-x)> that I wear"t comprehend.
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Sorry I simply understood this remains in the incorrect area, it is for a limitation concern however the concern itself is not. (I put on"t recognize just how to relocate this right into the best area)
According to Wolfram Alpha: \(\ displaystyle \ frac \) can likewise be created \(\ displaystyle \ frac -1 \)
Their solution is proper ... however actually stupid!I attempted to identify their thinking ... weird!We have:. \(\ displaystyle \ frac \)Multiply by \(\ displaystyle \ tfrac -1 \!:\;\; \ frac -1 -1 \ cdot \ frac \; =\; \ frac -x-y -x+y \; =\; \ frac \)Lengthy department:. \(\ displaystyle \ start selection ccccc & & & -1 \ \ & &-- &-- \ \ y-x &|& -y & -x \ \ & & -y & +x \ \ & &-- &-- \ \ & & & -2 x \ end range \)As a result:. \(\ displaystyle \ frac -y-x y-x \; =\; -1 - \ frac 2x \)|||||||||||||||||||||||| Why not separate "generally"?. \(\ displaystyle \ frac x+y \). \(\ displaystyle \ start selection & & & & 1 \ \ & &-- &-- &-- \ \ x-y &|& x & + & y \ \ & & x & -& y \ \ & &-- &-- &-- \ \ & & & & 2y \ end \)For that reason:. \(\ displaystyle \ frac \; =\; 1 + \ frac 2y \)
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