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Is x^2 greater than x-y?

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Intern
Joined: 28 Jun 2011
Posts: 13
Is x^2 greater than x-y?  [#permalink]

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13 May 2014, 06:31
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75% (hard)

Question Stats:

54% (01:57) correct 46% (01:30) wrong based on 131 sessions

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Is x^2 greater than x - y?

(1) y - x is positive
(2) x > 1 and y > 1

I have a questions about that problem. The answer for (1) says that y - x is positive so x- y is negative ((which equals -1(y-x))

How can I get a general rule for the highlighted statement? Thanks for the answers.
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Joined: 02 Sep 2009
Posts: 49493
Re: Is x^2 greater than x-y?  [#permalink]

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13 May 2014, 06:47
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steilbergauf wrote:
Is x^2 greater than x-y?

(1) y - x is positive
(2) x>1 and y>1

I have a questions about that problem. The answer for (1) says that y - x is positive so x- y is negative ((which equals -1(y-x))

How can I get a general rule for the highlighted statement? Thanks for the answers.

Is x^2 greater than x-y?

Is $$x^2>x-y$$? (The question can be re-phrased as is $$x^2-x>-y$$?)

(1) y - x is positive --> $$y-x>0$$ --> subtract y-x from both sides of the inequality:$$0>-(y-x)$$ --> $$0>x-y$$. Since $$x^2\geq{0}$$ (the square of any number is more than or equal to 0), then $$x^2\geq{0}>x-y$$. Sufficient.

(2) x>1 and y>1. Since $$x>1$$, then $$x^2-x>0$$ and since $$y$$ is positive, then $$-y$$ is negative. Thus, $$(x^2-x=positive)>(-y=negative)$$. Sufficient.

As for your question: we can get $$0>x-y$$ from $$y-x>0$$ by subtracting y-x from both sides of that inequality.

Hope it's clear.
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Re: Is x^2 greater than x-y?  [#permalink]

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12 Jun 2015, 17:57
Hi steilbergauf,

Since your original post is over a year old, I'm not sure if you're still around, but here's some extra information about the question you asked...

In Fact 1, we're told that Y - X is POSITIVE.

This means that Y > X. As a simple example, let's choose Y = 5, X = 2...

Thus, Y - X = 5 - 2 = 3

If you "flip" the order of the variables, then the larger number (Y) is subtracted from the smaller number (X), so the result MUST be NEGATIVE.

3 - 5 = -2

This concept can be taken a step further in the following example. Assuming X and Y are real numbers AND X is NOT equal to Y, what is the value of the following fraction...?

(X-Y)/(Y-X) = ?

Here's a hint: TEST VALUES (pick a value for X and Y and plug them in). What do you end up with? If you change the numbers, what happens to the answer?

Since (X-Y) and (Y-X) are exact opposites of one another (meaning positive and negative versions of the same number), you'll ALWAYS end up with -1.

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Re: Is x^2 greater than x-y?  [#permalink]

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13 Jan 2018, 09:58
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Re: Is x^2 greater than x-y? &nbs [#permalink] 13 Jan 2018, 09:58
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