How do you simplify #(x^2-y^2)/(8y-8x)#?

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Answer 1 · 2017-05-27T02:37:40

See a solution process below:

First, factor the numerator using this rule:

#a^2 - b^2 = (a + b)(a - b)#

#(x^2 - y^2)/(8y - 8x) => ((x + y)(x - y))/(8y - 8x)#

Next, factor a #color(red)(-8)# out of the denominator:

#((x + y)(x - y))/((-8 xx - y) + (-8 xx x)) => #

#((x + y)(x - y))/(-8(-y + x)) => #

#((x + y)(x - y))/(-8(x - y))#

Now, cancel the common terms in the numerator and the denominator:

#((x + y)color(red)(cancel(color(black)((x - y)))))/(-8color(red)(cancel(color(black)((x - y))))) =>#

#(x + y)/-8#

#-(x + y)/8#