# Showing that $\frac{x - \frac{x}{y}}{y - 1} = \frac{x}{y}$ by directly manipulating $\frac{x - \frac{x}{y}}{y - 1}$.

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 $$\frac{x}{y}$$?

• 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...) Sep 25 at 13:35
• Multiplying both sides by $y(y-1)$ is called clearing denominators. It's a common tactic for solving this sort of equation. Sep 25 at 13:48
• Also, don't forget that we need $y\neq 1$, and not only $y>0$ (are you sure, that $y<0$ is not allowed?) Sep 25 at 13:54
• I’m voting to close this question because it keeps attracting answers that all say the same. Sep 26 at 18:03

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.

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

$$\frac{x-\frac{x}{y}}{y-1}=\frac{\frac{xy-x}{y}}{y-1}=\frac{xy-x}{y}\cdot\frac{1}{y-1}=\frac{x(y-1)}{y}\cdot\frac{1}{y-1}=\frac{x}{y}$$

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!