If $0.8= \frac{y}{186+x}$, find the smallest value of $x$, if $x$ and $y$ are positive integers.
Note that $0.8 = \frac{8}{10} = \frac{4}{5}$, so $\frac{4}{5} = \frac{y}{186+x}$. As $x$ is positive, we want to find the smallest number greater than $186$ which is a multiple of $5$. This number is 190, which implies $\boxed{x=4}$.