This is a good question. The solution lies in good algebra technique.

z = x/y -- This expression shows z in terms of x and y.

Use multiplication to "solve for" x.

z * y = (x/y) * y = x * (y/y) = x*1 = x and we see that z * y = x -- This expression shows x in terms of y and z

Le's see you find an expression, using your best algebra, showing y in term of x and z.

Note: I deliberately didn't quite use the same thing you wrote. It's just an example of how it might be approached.