If $y\ne 0$ then the number of pairs $(x,y)$ satisfying $x+y+\frac{x}{y}=\frac{1}{2}, (x+y)\left(\frac{x}{y}\right)=-2$ is 

$x+y=v;\frac{x}{y}=\frac{-2}{v}$

$\therefore v-\frac{2}{v}=\frac{1}{2}$

$2v^2-v-4=0$

$v=\frac{1\pm \sqrt{33}}{4}$

$x+y=\frac{1+\sqrt{33}}{4}\Rightarrow \frac{x}{y}=\frac{1-\sqrt{33}}{4}\&\text{ x+y=}\frac{1-\sqrt{33}}{4}\Rightarrow y=\frac{1-\sqrt{33}}{5+\sqrt{33}}$

$x=\frac{-8}{5+\sqrt{33}}$

$(x,y)=(\frac{-8}{5+\sqrt{33}},\frac{1-\sqrt{33}}{5+\sqrt{33}})or\left(\frac{8}{\sqrt{33-5}}\frac{1+\sqrt{33}}{5-\sqrt{33}}\right)$