What is #(x^3y-2xy+xy^2) ÷xy#?

1 Answer
Alippiun
Apr 14, 2016

#x^2-2+y#

Explanation:

Firstly express the calculation in as fraction ;

#(x^3y-2xy+xy^2)-:xy#

#=(x^3y-2xy+xy^2)/(xy)#

There are only one variable in the denominator, so we can express one whole fraction into smaller fraction , but with the same denominator and solve for each of it;

#=(x^3y-2xy+xy^2)/(xy)#

#=(x^(cancel(3)2)cancely)/(cancel(xy))-(2cancel(xy))/(cancel(xy))+(cancelxy^cancel(2))/(cancel(xy))#

#=x^2-2+y#

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