SOLUTION: x^2n + 2x^n y^n - 3y^2n over x^2n + 5x^n y^n - 6y^2n on the expressions. the 2n is the exponent to the x. and on the 2x, then n is the exponent, and it is multiplies with the y^

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: x^2n + 2x^n y^n - 3y^2n over x^2n + 5x^n y^n - 6y^2n on the expressions. the 2n is the exponent to the x. and on the 2x, then n is the exponent, and it is multiplies with the y^      Log On


   

Question 173730This question is from textbook
: x^2n + 2x^n y^n - 3y^2n over x^2n + 5x^n y^n - 6y^2n
on the expressions. the 2n is the exponent to the x. and on the 2x, then n is the exponent, and it is multiplies with the y^n. It is the same with the next expression
The directions are just to simplify.
Thank you! :) This question is from textbook

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


This looks like a horror, but it really isn't that bad.

Let t=x%5En and u=y%5En then substitute:

%28t%5E2+%2B+2tu+-+3u%5E2%29%2F%28+t%5E2+%2B+5tu+-+6u%5E2%29

Now factor both the numerator and the denominator:

%28%28t-u%29%28t%2B3u%29%29%2F%28%28t-u%29%28t%2B6u%29%29

Eliminate the common factor of t-u

%28cross%28%28t-u%29%29%28t%2B3u%29%29%2F%28cross%28%28t-u%29%29%28t%2B6u%29%29

Then substitute the original expressions for t and u back into the expression:

%28x%5En%2B3y%5En%29%2F%28x%5En%2B6y%5En%29

Done.