Suppose that x and y are nonzero real numbers such that (3x+y)/(x-3y)=-2. What is the value of (x+3y)/(3x-y)?

2017 AMC 10B Problem 4

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Suppose 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}?

-3

-1

1

2

3

Clear the denominator in the given equation

The equation quickly forces a relation between

x

and

y

Given that

\frac{3x+y}{x-3y}=-2,

we can multiply by the denominator to get

3x+y = 6y-2x.

Solving, we can see that

x=y.

Therefore,

\frac{x+3y}{3x-y} = \frac{x+3x}{3x-x}=2.

Thus, the correct answer is D .