How do you express #((y^2-xy)/(x^2+xy)-xy+y^2)(x/(x-y))+(y/(x+y))# as a single fraction?

1 Answer
Shwetank Mauria
May 26, 2017

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

Explanation:

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

#=((-y(x-y))/(x(x+y))-y(x-y))(x/(x-y))+(y/(x+y))#

#=((-y(x-y))/(x(x+y))+(-y(x-y)x(x+y))/(x(x+y)))(x/(x-y))+(y/(x+y))#

#=((-y(x-y)(1+x(x+y)))/(x(x+y)))(x/(x-y))+(y/(x+y))#

#=((-ycancel((x-y))(1+x(x+y)))/(cancelx(x+y)))(cancelx/cancel((x-y)))+(y/(x+y))#

#=(-y(1+x(x+y)))/((x+y))+(y/(x+y))#

#=-y/((x+y))-(yx(x+y))/((x+y))+(y/(x+y))#

#=-xy#

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