How do you simplify $\frac { x ^ { 2} - x y - ( x y - y ^ { 2} ) } { ( x ^ { 2} - y ^ { 2} ) }$ ?

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Answer 1 · 2017-09-18T10:33:28

$(x-y)/(x+y)$

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

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

$= (x-y)^cancel2/((x+y)cancel((x-y))) = (x-y)/(x+y)$ [Ans]

Answer 2 · 2017-09-18T10:43:42

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

Let us begin by simplifying the numerator:

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

Adding like terms,

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

Notice the identities in the numerator and denominator.
Expressions of the form:

$a^2 - 2ab +b^2 = (a-b)(a-b)=(a-b)^2$
And
$a^2 - b^2 = (a+b)(a-b)$

Therefore,
$x^2 - 2xy +y^2= (x-y)(x-y)=(x-y)^2$
And
$x^2 - y^2 = (x+y)(x-y)$

Let us rewrite the expression now, using the identities:
$(x^2 - 2xy + y^2)/(x^2 - y^2) = ((x-y)(x-y))/((x+y)(x-y))$

Then cancel out the common $(x-y)$ in both the numerator and denominator:

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

How do you simplify \frac { x ^ { 2} - x y - ( x y - y ^ { 2} ) } { ( x ^ { 2} - y ^ { 2} ) }\frac { x ^ { 2} - x y - ( x y - y ^ { 2} ) } { ( x ^ { 2} - y ^ { 2} ) }?