KhanAcademy nested fractions problem

Question

So I was given a problem at KA today. They offer you a choice among possible simplified versions of this expression:

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

My solution was:

$$(1+\frac{x}{y})\div\frac{x}{y}=(1+\frac{x}{y})\cdot\frac{y}{x}=\frac{y}{x}+\frac{\not x}{\not y}\cdot\frac{\not y}{\not x}=\frac{y}{x}$$

And it was one of the variants, but I was told this was wrong. They solve this with a different, perfectly reasonable strategy (they get $\frac{y+x}{x}$), but I still don't get what is wrong with my approach.

Answer 1

When you cancel out $\frac{x}{y}$ with $\frac{y}{x}$, it doesn't give you $0$, it gives you $1$.

For instance, $\frac{3}{4}\cdot\frac{4}{3} = 1$.

Answer 2

Your solution is almost correct note that $$\frac{1 + \frac{x}{y}}{\frac{x}{y}} = \frac{y}{x} \left( 1 + \frac{x}{y} \right) = \frac{y}{x} + 1$$ The cancelling out does not make the fraction $0$!