Rearranging, we find $5 x+y=-x+4y$, or $6x=3 y \Longrightarrow y=2x$. Substituting, we can convert the second equation into $$\frac{x+6 x}{3 x-2x}=\frac{7 x}{x}=\text { (C) } 7$$. 
Suppose that $x$ and $y$ are nonzero real numbers such that $\frac{5 x+y}{x-4y}=-1$. What is the value of $\frac{x+3 y}{3 x-y}$? (Adapted From 2017 AMC 10B Problems, Question \#4) 
Options:
(1) $$2$$ 
(2) $$5$$ 
(3) $$7$$ 
(4) $$9$$ 
(5) $$11$$ 