The short answer is: They are not exactly equivalent.
Consider the following system:
$$
\begin{cases}
\dot x=X(x,y)\\
\dot y=Y(x,y)
\end{cases}
$$
and the first order differential equation
$$
\frac{dy}{dx}=\frac{Y(x,y)}{X(x,y)}.\
$$
These are equivalent in a neighborhood of a point $(x_0,y_0)$ if $X(x_0,y_0)\neq 0$ in the following sense: the integral curves (the graphs of the solutions) of the first order equation are exactly the phase curves (images of the solutions of the system) of our planar system in this this neighborhood. If at the point $(x_0,y_0)$ $X(x_0,y_0)=0$ but $Y(x_0,y_0)\neq 0$ then the phase curves will correspond to the integral curves of
$$
\frac{dx}{dy}=\frac{X(x,y)}{Y(x,y)}.\
$$
And the problems start when both $X(x_0,y_0)=0$ and $Y(x_0,y_0)=0$ then the phase portrait has an equilibrium point there, but integral curves are simply not defined at this particular point.