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Supposed that $x$ and $y$ are nonzero real numbers such that $\frac{3 x+y}{x-3 y}=-2$. What is the value of $\frac{x+3 y}{3 x-y}$? 
Overseas Competition->Knowledge Point->Calculation Modules->Basic Concepts of Equation->Proportional Equations
Rearranging, we find $3 x+y=-2 x+6 y$, or $5 x=5 y \Longrightarrow x=y$. Substituting, we can convert the second equation into $\frac{x+3 x}{3 x-x}=\frac{4 x}{2 x}= 2$ More step-by-step explanation:  $$ \begin{aligned} \&\frac{3 x+y}{x-3 y}=-2 \textbackslash\textbackslash{} \&3 x+y=-2(x-3 y) \textbackslash\textbackslash{} \&3 x+y=-2 x+6 y \textbackslash\textbackslash{} \&5 x=5 y \textbackslash\textbackslash{} \&x=y \textbackslash\textbackslash{} \&\frac{x+3 y}{3 x-y}=\frac{1+3(1)}{3(1)-1}=\frac{4}{2}=2 \end{aligned} $$ 