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# m18 08

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07 Sep 2010, 17:19
i'm not sure how to solve this problem.

If neither [m]x[/m] nor [m]y[/m] is 0, what is [m]\frac{2x + y}{x - 2y}[/m] ?

1. [m]\frac{x}{x + y} = 2[/m]
2. [m]x - 2y = 4[/m]
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Re: m18 08 [#permalink]

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07 Sep 2010, 17:35
woodahy wrote:
i'm not sure how to solve this problem.

If neither $$x$$ nor $$y$$ is 0, what is $$\frac{2x + y}{x - 2y}$$ ?

1. $$\frac{x}{x + y} = 2$$
2. $$x - 2y = 4$$

(1) $$\frac{x}{x + y} = 2$$ --> $$x=2x+2y$$ --> $$x=-2y$$ --> $$\frac{2x + y}{x - 2y}=\frac{-4y+y}{-2y-2y}=\frac{-3y}{-4y}=\frac{3}{4}$$. Sufficient.

(2) $$x - 2y = 4$$ --> $$x=2y+4$$ --> $$\frac{2x + y}{x - 2y}=\frac{2y+4+y}{4}=\frac{3y+4}{4}$$ --> we don't know the value of $$y$$, hence we can not calculate numerical value of this fraction. Not sufficient.

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Re: m18 08 [#permalink]

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07 Sep 2010, 17:52
Thanks. Much better than explanation given. I had 3y/4y and didn't realize I had the answer

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Re: m18 08   [#permalink] 07 Sep 2010, 17:52
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# m18 08

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