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Is x^2 greater than x-y?

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

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New post 13 May 2014, 06:31
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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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Re: Is x^2 greater than x-y?  [#permalink]

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New post 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.

Answer: D.

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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New post 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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New post 01 Oct 2019, 06:49
steilbergauf wrote:
Is \(x^2\) greater than x - y?

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


\(x^2>x-y…x^2-x>-y…y<-x^2+x…y<x-x^2…y<x(1-x)?\)

(1) y - x is positive: sufic.
if \(y-x>0…y>x\) then is \(x<x-x^2…0<x-x^2-x…0<-x^2…0>x^2?\)
any \(number^2≥0\) so \(0>x^2\) is always false

(2) x > 1 and y > 1: sufic.
if \(x,y>1\) then \(x(1-x)=negative\), so is \(y<x(1-x)…y<negative?\)
since \(y>1\) then \(y<negative\) is false

Answer (D)
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Re: Is x^2 greater than x-y?   [#permalink] 01 Oct 2019, 06:49
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