How do you simplify $\frac{x^{2} - y^{2}}{8 y - 8 x}$ $(x^2-y^2)/(8y-8x)$ ?

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Answer 1 · 2017-05-27T02:37:40

See a solution process below:

First, factor the numerator using this rule:

$a^{2} - b^{2} = (a + b) (a - b)$ $a^2 - b^2 = (a + b)(a - b)$

$\frac{x^{2} - y^{2}}{8 y - 8 x} \Rightarrow \frac{(x + y) (x - y)}{8 y - 8 x}$ $(x^2 - y^2)/(8y - 8x) => ((x + y)(x - y))/(8y - 8x)$

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

$\frac{(x + y) (x - y)}{(- 8 \times - y) + (- 8 \times x)} \Rightarrow$ $((x + y)(x - y))/((-8 xx - y) + (-8 xx x)) =>$

$\frac{(x + y) (x - y)}{- 8 (- y + x)} \Rightarrow$ $((x + y)(x - y))/(-8(-y + x)) =>$

$\frac{(x + y) (x - y)}{- 8 (x - y)}$ $((x + y)(x - y))/(-8(x - y))$

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

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

$(x + y)/-8$

$-(x + y)/8$

How do you simplify (x^2-y^2)/(8y-8x)x2−y28y−8x(x^2-y^2)/(8y-8x)?