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

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:


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.

  • 1
    $\begingroup$ $x-y=-(y-x)$.${}$ $\endgroup$ – David Mitra Jan 31 '14 at 15:41
  • 1
    $\begingroup$ $\dfrac{y-x}{x-y} = -1$. 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

You did fine work up to where you're "stuck": Just note that $y - x = -(-y + x) = -(x-y)$: $\require{cancel}$ $$ {(y-x)(y+x) \over{x-y}} = \dfrac{-(\cancel{x-y})(y+x)}{\cancel{x-y}} = - (y+x) = -(x + y)$$

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.