Whats the simplified form of #(1/y-1/x)/(1/y+1/x)#?

1 Answer
Jun 9, 2017

See a solution process below:

Explanation:

To be able to add and subtract the fractions in the numerator and denominator of the main term we need to have each fraction over a common denominator.

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

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

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

We can now use this rule for dividing fractions to simplify the expression:

#(color(red)(a)/color(blue)(b))/(color(green)(c)/color(purple)(d)) = (color(red)(a) xx color(purple)(d))/(color(blue)(b) xx color(green)(c))#

#(color(red)(x - y)/color(blue)(xy))/(color(green)(x + y)/color(purple)(xy)) = ((color(red)(x - y)) xx color(purple)(xy))/(color(blue)(xy) xx (color(green)(x + y))) = ((color(red)(x - y)) xx cancel(color(purple)(xy)))/(cancel(color(blue)(xy)) xx (color(green)(x + y))) =#

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