what we can do is solve both equations for "y", so they end up in slope-intercept form, noticing that the second equation is already in slope-intercept form.
now, keep in mind that parallel lines, have the same slope, and perpendicular ones have a negative reciprocal slope.
[tex]\bf \begin{cases}
-5y=2x+3\implies y=\cfrac{2x+3}{-5}\implies &\boxed{y=\stackrel{slope}{-\cfrac{2}{5}}x-\cfrac{3}{5}}\\\\
y=\cfrac{2}{4}x+4\implies &\boxed{y=\stackrel{slope}{\cfrac{1}{2}}x+4}
\end{cases}[/tex]
notice, the slopes differ, so clearly they're not parallel.
are they perpendicular? well, the product of the perpendicular ones gives -1, let's check, [tex]\bf -\cfrac{2}{5}\cdot \cfrac{1}{2}\implies -\cfrac{1}{5}[/tex]
nope, no -1 as product, thus they're not perpendicular either, so they're neither then.
