2017 AMC 10 B

Complete problem set with solutions and individual problem pages

Problem 4 Easy

Suppose that x and y are nonzero real numbers such that \dfrac{3x+y}{x-3y}=-2. What is the value of \dfrac{x+3y}{3x-y}? (2017 AMC 10B Problem, Question#4)

  • A.

    -3

  • B.

    -1

  • C.

    1

  • D.

    2

  • E.

    3

Answer:D

Rearranging, we find 3x+y=-2x+6y, or 5x=5y\Rightarrow x=y. Substituting, we can convert the second equation into \dfrac{x+3x}{3x-x}=\dfrac{4x}{2x}=2.

Substituting each x and y with 1, we see that the given equation holds true, as \dfrac{3\left( 1\right)+1}{1-3\left( 1\right)}=-2. Thus,\dfrac{x+3x}{3x-x}=-2.

Let y=ax. The first equation converts into \dfrac{\left( 3+a\right)x}{\left( 1-3a\right)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}=2.

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