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 $$\frac{x+3 x}{3 x-x}=\frac{4 x}{2 x}=\text { (D) } 2$$. 
Supposed that $x$ and $y$ are nonzero real numbers such that $\frac{3 x+y}{x-3 y}=-2$. What is the value of $\frac{x+3 y}{3 x-y}$? (2017 AMC 10B Problems, Question \#4) 

a: $$-3$$ 
b: $$-1$$ 
c: $$1$$ 
d: $$2$$ 
e: $$3$$ 