I am trying to simplify

$$\frac{\frac{y}{x} - \frac{x}{y}}{\frac{1}{y} - \frac{1}{x}}$$

I make the top part into

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

I know the bottom can be rewritten to just be multiplied into the top as its inverse.

$$\frac{y^2 - x ^2}{xy} * (y - x)$$

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

This of course is wrong, but I do not know why, it appears to be correct to me. Nothing I did is mathematically wrong, it follows all the rules. What went wrong?


3 Answers 3


The inverse of the denominator $$\frac{1}{y}-\frac{1}{x}$$ is not $y-x$. The denominator is equal to $$\frac{1}{y}-\frac{1}{x}=\frac{x-y}{xy},$$ so its inverse is $$\frac{xy}{x-y},$$ so the entire fraction is $$\frac{\frac{y}{x}-\frac{x}{y}}{\frac{1}{y}-\frac{1}{x}}=\frac{\frac{y^2-x^2}{xy}}{\frac{1}{y}-\frac{1}{x}}=\frac{\frac{y^2-x^2}{xy}}{\frac{x-y}{xy}}=\frac{y^2-x^2}{xy}\cdot\frac{xy}{x-y}=\frac{y^2-x^2}{x-y}$$ There's one more simplification you can make at this step; do you see it?


No, you didn’t follow the rules: there is nothing in the rules that lets you transform


into $y-x$, which is what you’ve done when you go from




You can’t invert the denominator until you’ve transformed it into a single fraction:


Now you can invert and multiply:


  • $\begingroup$ I don't follow, I never put a 1 ontop. $\endgroup$
    – user138246
    Commented May 18, 2013 at 18:29
  • 1
    $\begingroup$ @Jordan: It’s there implicitly: for any quantities $a$ and $b$, the fraction $\frac{a}b$ is the same thing as $a\cdot\frac1b$. When $b=\frac{c}d$, this means that $\frac{a}b=a\cdot\frac1{c/d}$, and the invert and multiply rule works only because $\frac1{c/d}=\frac{d}c$. You can go from multiplying $a$ by $\frac1{c/d}$ to multiplying $a$ by $\frac{d}c$ because you’re multiplying by the same thing in both cases. You went from multiplying by $\frac1{\frac17-\frac1x}$ to multiplying by $y-x$, which doesn’t work, because these two quantities are not equal. $\endgroup$ Commented May 18, 2013 at 18:37
  • $\begingroup$ It would have worked if it was $\frac{1}{y-x}$ right? $\endgroup$
    – user138246
    Commented May 18, 2013 at 18:41
  • 2
    $\begingroup$ @Jordan: You mean if the denominator of the original monster fraction had been $\frac1{y-x}$? Yes, absolutely. $\endgroup$ Commented May 18, 2013 at 18:42



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