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 ?