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

Question

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 $\frac{x}{y}$?

Answer 1

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.

Answer 2

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

Answer 3

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 equationwe 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!