How do you simplify #((x/y)-(y/x))/((1/y)-(1/x))#?

1 Answer
George C.
Apr 20, 2017

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

with exclusions #x!=0#, #y!=0#, #x!=y#.

Explanation:

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

#color(white)((x/y-y/x)/(1/y-1/x)) = (x^2-y^2)/(x-y)#

#color(white)((x/y-y/x)/(1/y-1/x)) = (color(red)(cancel(color(black)((x-y))))(x+y))/(color(red)(cancel(color(black)(x-y)))#

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

with exclusions #x!=0#, #y!=0#, #x!=y#.

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