# Answer to Question #24149 in Abstract Algebra for Xane

Question #24149

How to do this sum: 2 over x-y plus 1 over x( to the power of two) - y( to the power of two)

Expert's answer

We have the following statement:

A = 2/(x-y) + 1/(x²-y²).

First, let's reduce this statement to the common denominator. The common denominator is

(x²-y²),

as

(x²-y²) = (x+y)(x-y).

Therefore

A = 2/(x-y) + 1/(x²-y²) =

= 2(x+y)/((x-y)(x+y)) + 1/(x²-y²) =

= 2(x+y)/(x²-y²) + 1/(x²-y²) =

= (2(x+y) + 1)/(x²-y²) =

= (2x+2y+1)/(x²-y²).

So,

2/(x-y) + 1/(x²-y²) = (2x+2y+1)/(x²-y²).

A = 2/(x-y) + 1/(x²-y²).

First, let's reduce this statement to the common denominator. The common denominator is

(x²-y²),

as

(x²-y²) = (x+y)(x-y).

Therefore

A = 2/(x-y) + 1/(x²-y²) =

= 2(x+y)/((x-y)(x+y)) + 1/(x²-y²) =

= 2(x+y)/(x²-y²) + 1/(x²-y²) =

= (2(x+y) + 1)/(x²-y²) =

= (2x+2y+1)/(x²-y²).

So,

2/(x-y) + 1/(x²-y²) = (2x+2y+1)/(x²-y²).

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