Uhm, you just said it, y = 2

But to solve $\displaystyle \frac{x}{y} = 5$, you can rearrange as follows :

>> For $\displaystyle x$ :

$\displaystyle \frac{x}{y} = 5$

$\displaystyle \frac{x}{y} \times y = 5 \times y$

$\displaystyle \frac{x \times y}{y} = 5 \times y$

$\displaystyle \frac{x}{1} = 5 \times y$ (cancel out the $\displaystyle y$)

$\displaystyle x = 5 \times y$

$\displaystyle x = 5y$

>> For $\displaystyle y$ :

$\displaystyle \frac{x}{y} = 5$

$\displaystyle \frac{y}{x} = \frac{1}{5}$ (inverse on both sides)

$\displaystyle \frac{y}{x} \times x = \frac{1}{5} \times x$

$\displaystyle \frac{xy}{x} = \frac{x}{5}$

$\displaystyle \frac{y}{1} = \frac{x}{5}$ (cancelling out)

$\displaystyle y = \frac{x}{5}$

$\displaystyle y = \frac{1}{5} x$

This is how I would do it (the long way of course, in real situations I would spare most steps). Does it make sense ?