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$$\frac yx-\frac xy \over \frac 1y- \frac 1x$$

I am trying to solve this, I start with the top, multiply the left side by Y and the right side by x to get:

$$\frac{y^2-x^2}{xy}$$

then I go to the bottom and multiply the left side by x and the right side by y to get:

$$\frac{y-x}{xy}$$ so that gives me:

$${(y-x)(y+x) \over xy} \over{x-y \over xy}$$

I then multiply the top by the inverse of the bottom, which should cancel out the xy and then I am left with :

$$(y-x)(y+x) \over{x-y}$$

I assume I should factor out a -1 from the bottom but the answer states "-(x+y)"

I'm missing something.

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    $\begingroup$ $x-y=-(y-x)$.${}$ $\endgroup$ – David Mitra Jan 31 '14 at 15:41
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    $\begingroup$ $\dfrac{y-x}{x-y} = -1$. Multiply with $y+x = x+y$. $\endgroup$ – Daniel Fischer Jan 31 '14 at 15:41
  • $\begingroup$ Thanks, I'm sorry about this. Small details escape me sometimes. $\endgroup$ – Joshhw Jan 31 '14 at 15:53
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You did fine work up to where you're "stuck": Just note that $y - x = -(-y + x) = -(x-y)$: $\require{cancel}$ $$ {(y-x)(y+x) \over{x-y}} = \dfrac{-(\cancel{x-y})(y+x)}{\cancel{x-y}} = - (y+x) = -(x + y)$$

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  • $\begingroup$ Diagonal cancelling lines, neat! Didn't know about those. $\endgroup$ – Dahn Jan 31 '14 at 16:05

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