As Exponents

$\begin{aligned} \frac {2^x 5^y}{10^9} & = 8 \\ 2^x 5^y & = 8 \times 10^9 = 2^{12} \times 5^9 \end{aligned}$

$\implies \begin{cases} x = 12 \\ y = 9 \end{cases}$ .

Note that $x=12$ and $y=9$ also satisfy the second equation $\dfrac {2^9 \times 5^{12}}{10^9} = 5^3 = 125$ . Therefore, $\dfrac yx = \dfrac {12}9 = \dfrac 43 = \boxed{0.75}$

Since $2$ and $5$ are prime factors of $10$ , the expressions above can be converted to $2^{x-9}5^{y-9}$ and $2^{y-9}5^{x-9}$ respectively, with the exponent of $10$ subtracting from the exponents of $2$ and $5$ .

$8$ has a prime factor of $2$ . This means that in the equation $2^{x-9}5^{y-9}=8$ , $5^{y-9}$ should equal to $1$ .

$\large \begin{aligned} 5^{y-9} &= 1 \\ 5^{9-9} &= 1 \\ 5^{0} &= 1 \\ \implies y &= 9 \end{aligned}$

This leaves us with $2^{x-9}=8$ . Since $8=2^{3}$ :

$\large \begin{aligned} 2^{x-9} &= 8 \\ 2^{x-9} &= 2^{3} \\ x-9 &= 3 \\ x &= 12 \end{aligned}$

Getting the values of $x$ and $y$ , we can now solve $\frac{y}{x}$ which is $\frac{9}{12}$ or $\boxed{0.75}$ .

This method can also be done to $2^{y-9}5^{x-9}=125$ or the second equation of the problem, with $2^{y-9}$ equalling to $1$ since $125$ has $5$ as its only prime factor.