$\frac{x}{y}+\frac{y}{x}=-2\Rightarrow \frac{{{x}^{2}}+{{y}^{2}}}{xy}=-2\Rightarrow {{x}^{2}}+{{y}^{2}}+2xy=0\Rightarrow {{(x+y)}^{2}}=0$

در نتیجه: $x+y=0\Rightarrow y=-x\,\,\,\,\,\,\,\,\,\,\,\,\,x\ne 0\,\,\,\,,\,\,\,\,y\ne 0$
بنابراین y تابعی از x است.

$\left\{ \begin{matrix}
   y+2y+3=x\,\,\,\,\,\,\,\,\,y\ge -\frac{3}{2}  \\
   y-2y-3=x\,\,\,\,\,\,\,\,\,y<-\frac{3}{2}  \\
\end{matrix}\Rightarrow \left\{ \begin{matrix}
   y=\frac{1}{3}x-1\,\,\,\,\,\,y\ge -\frac{3}{2}  \\
   y=-x-3\,\,\,\,\,\,y\lt -\frac{3}{2}  \\
\end{matrix} \right. \right.$
$x=0\Rightarrow \left\{ \begin{matrix}
   y=-1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y\ge -\frac{3}{2}\,\,Tabe\,Nist  \\
   y=-3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y<-\frac{3}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,  \\
\end{matrix} \right.$
$\left\{ \begin{matrix}
   x+\sqrt{y+4}=y+2  \\
   x=-2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,  \\
\end{matrix}\Rightarrow -2+\sqrt{y+4}=y+2\Rightarrow \sqrt{y+4}=y+4 \right.$

در این صورت:

$\begin{align}
  & y+4=0\Rightarrow y=-4\,Tabe\,Nist \\
 & y+4=1\Rightarrow y=-3 \\
 & y+x=y+3\Rightarrow x=3\,Tabe\,Nist \\
\end{align}$