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

Question

2462 views around the world

You can reuse this answer
Creative Commons License

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)#