# 2017 AMC 10B Problems/Problem 4

## Problem

Supposed that $x$ and $y$ are nonzero real numbers such that $\frac{3x+y}{x-3y}=-2$. What is the value of $\frac{x+3y}{3x-y}$? $\textbf{(A)}\ -3\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 3$

## Solution 1

Rearranging, we find $3x+y=-2x+6y$, or $5x=5y\implies x=y$. Substituting, we can convert the second equation into $\frac{x+3x}{3x-x}=\frac{4x}{2x}=\boxed{\textbf{(D)}\ 2}$.

More step-by-step explanation: $\frac{3x+y}{x-3y}=-2$ $3x+y=-2\left(x-3y\right)$ $3x+y=-2x+6y$ $5x=5y$ $x=y$ $\frac{x+3y}{3x-y}=\frac{1+3\left(1\right)}{3\left(1\right)-1}=\frac{4}{2}=2$.

We choose $\boxed{\textbf{(D)}\ 2}$.

## Solution 2

Substituting each $x$ and $y$ with $1$, we see that the given equation holds true, as $\frac{3(1)+1}{1-3(1)} = -2$. Thus, $\frac{x+3y}{3x-y}=\boxed{\textbf{(D)}\ 2}$

## Solution 3

Let $y=ax$. The first equation converts into $\frac{(3+a)x}{(1-3a)x}=-2$, which simplifies to $3+a=-2(1-3a)$. After a bit of algebra we found out $a=1$, which means that $x=y$. Substituting $y=x$ into the second equation it becomes $\frac{4x}{2x}=\boxed{\textbf{(D)}\ 2}$ - mathleticguyyy

## Solution 4

Let $x=1$. Then $y=1$. So the desired result is $2$. Select $\boxed{D}$.

~hastapasta

~ pi_is_3.14

~savannahsolver

## Video Solution by TheBeautyofMath

~IceMatrix

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