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As I was trying to help my niece with her homework, I realized that the only way I know how to conclude why $\frac{x - \frac{x}{y}}{y - 1} = \frac{x}{y}$ for $x, y > 0$ is by multiplying both sides of the equation by $y(y-1)$. But this is a bit unsatisfactory if you would like to "discover" the possible equality of $\frac{x - \frac{x}{y}}{y - 1}$ without any prior knowledge of what it might be. How should one manipulate $\frac{x - \frac{x}{y}}{y - 1}$ directly to obtain $\frac{x}{y}$?

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    $\begingroup$ Focus on the numerator only, you get $x-\frac{x}{y}=\frac{xy-x}{y}=\frac{x(y-1)}{y}$. So, dividing by $y-1$ gives $\frac{x}{y}$. This is a pretty natural thing to do because the numerator looks complicated, so why not start off by simplifying it. (You also need to exclude $y=1$, but you mentioned niece's hw so...) $\endgroup$
    – peek-a-boo
    Sep 25 at 13:35
  • $\begingroup$ Multiplying both sides by $y(y-1)$ is called clearing denominators. It's a common tactic for solving this sort of equation. $\endgroup$ Sep 25 at 13:48
  • $\begingroup$ Also, don't forget that we need $y\neq 1$, and not only $y>0$ (are you sure, that $y<0$ is not allowed?) $\endgroup$ Sep 25 at 13:54
  • $\begingroup$ I’m voting to close this question because it keeps attracting answers that all say the same. $\endgroup$
    – Kurt G.
    Sep 26 at 18:03

3 Answers 3

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First factor out $x$ and then $\frac{1}{y}$ leads us to\begin{align*} \frac{x - \frac{x}{y}}{y - 1} &= \frac{x(1 - \frac{1}{y})}{y - 1} \\ &= \frac{\frac{x}{y}(y - 1)}{y-1} = \frac{x}{y}\end{align*}

OFC one could factor our $\frac{x}{y}$ directly.

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    $\begingroup$ $\dfrac{x - \frac{x}{y}}{y - 1} = \dfrac{x(1 - \frac{1}{y})}{y - 1} = \dfrac{{x}(1 - \frac{1}{y})}{y(1 - \frac{1}{y})} = \dfrac{x}{y}$ works too $\endgroup$
    – Henry
    Sep 25 at 15:30
  • $\begingroup$ @Henry Thanks for the hint… Honestly, I like your way a bit more then mine :D $\endgroup$
    – Gono
    Sep 25 at 18:09
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$$ \frac{x-\frac{x}{y}}{y-1}=\frac{\frac{xy-x}{y}}{y-1}=\frac{xy-x}{y}\cdot\frac{1}{y-1}=\frac{x(y-1)}{y}\cdot\frac{1}{y-1}=\frac{x}{y} $$

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Assuming that you have already explained her what it actually means for something to be equal and connection of rational numbers to this $\frac{x}{y}$ and $ \frac{x-\frac{x}{y}}{y-1}$.

${x-\frac{x}{y}}$ and ${y-1}$ are both written in the $\frac{p}{q}$ form. In which $p={x-\frac{x}{y}}$ in which $y\neq0$ and $q=y-1$ in which y $\neq+1$.

Using this knowledge we can manipulate the RHS to get the desired result.

$\frac{x-\frac{x}{y}}{y-1}=({\frac{x}{1}-\frac{x}{y}})/(y-1)$ 🍀

Combining the fractions, ${\frac{x}{1}-\frac{x}{y}}\rightarrow \frac{yx-x}{y}$. Factoring out $x$ then $\frac{x(y-1)}{y}$ so this is our $p.$

Back to our original equation🍀we can rewrite it as,

$\frac{x(y-1)}{y-1}=(\frac{x(y-1)}{y})/(y-1)$.

Note that division is just multiplicative inverse meaning $\frac{x(y-1)}{y}\cdot\frac{1}{(y-1)}=\frac{x}{y}$ as desired!

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