# SOLUTION: (y^-5 - 3y^5x^-2 + y^5) x^-2 ________________________________________ (x^2 - p + x^-2) y^5 On this problem I am supposed to use the distributive p

Algebra ->  Algebra  -> Distributive-associative-commutative-properties -> SOLUTION: (y^-5 - 3y^5x^-2 + y^5) x^-2 ________________________________________ (x^2 - p + x^-2) y^5 On this problem I am supposed to use the distributive p      Log On

 Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help! Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations!

 Algebra: Distributive, associative, commutative properties, FOIL Solvers Lessons Answers archive Quiz In Depth

 Question 66818This question is from textbook Algebra 1 : (y^-5 - 3y^5x^-2 + y^5) x^-2 ________________________________________ (x^2 - p + x^-2) y^5 On this problem I am supposed to use the distributive property and write the answer with all exponents positive. I think I would multiply x^-2 by everything on the top and multiply y^5 by everything on the bottom. Where I get confused is changing all exponents to positive. If they're negative I move them to the opposite side and change the sign, right? Somehow, I'm just not getting the right answer. Thanks.This question is from textbook Algebra 1 Answer by stanbon(57219)   (Show Source): You can put this solution on YOUR website!(y^-5 - 3y^5x^-2 + y^5) x^-2 ________________________________________ (x^2 - p + x^-2) y^5 --------------------------------- Put that x^-2 in the denominator and multiply thru by it to get: (1/y^5 - 3y^5/x^2 + y^5) _________________________________ (x^4 - x^2p + 1)y^5 There is not much more you can do with this except distribute in the denominator to get: [(x^2 - 3y^10 + x^2y^10) ________________________________ (x^4 -x^2p + 1)x^2y^10 [(x^2 - 3y^10 + x^2y^10) _________________________________ [x^6y^10 - x^4py^10 +x^2y^10] I'm sure you could stop at a different point but this is as good as any. It's a mess of a problem. Cheers, Stan H.