How do you add or subtract #(4xy)/(x^2-y^2) + ( x-y)/(x+y)#?

1 Answer
Jun 28, 2018

See a solution process below:

Explanation:

To add or subtract fractions they must be over a common denominator.

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

Therefore we need to multiply the fraction on the right by #(x - y)(x - y)#, a form of #1#, to get both fractions over a common denominator:

#(4xy)/(x^2 - y^2) + ((x - y)/(x - y) xx (x - y)/(x + y)) =>#

#(4xy)/(x^2 - y^2) + ((x - y)(x - y))/((x - y)(x + y)) =>#

#(4xy)/(x^2 - y^2) + (x^2 - 2xy + y^2)/(x^2 - y^2)#

Now, we can add the numerators of the two fractions over their common denominator:

#(4xy + x^2 - 2xy + y^2)/(x^2 - y^2) =>#

#(x^2 + 4xy - 2xy + y^2)/(x^2 - y^2) =>#

#(x^2 + (4 - 2)xy + y^2)/(x^2 - y^2) =>#

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

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

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

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