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Factoring...? (m02 #8)

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Factoring...? (m02 #8) [#permalink] New post 27 Feb 2008, 21:50
There is this four step factor... and I just don't get step #2.

x(x-1) > y(y-4)
(y+1)(y+1-1) > y(y-4)
y+1 > y-4
1 > 4
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Re: Factoring...? (m02 #8) [#permalink] New post 27 Feb 2008, 23:59
looks like X has been substituted by y+1
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Re: Factoring...? (m02 #8) [#permalink] New post 28 Feb 2008, 01:18
Liquid wrote:
There is this four step factor... and I just don't get step #2.

x(x-1) > y(y-4)
(y+1)(y+1-1) > y(y-4)
y+1 > y-4
1 > 4


post the question in detail.
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Re: Factoring...? (m02 #8) [#permalink] New post 28 Feb 2008, 19:59
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The question is:


x any y are both positive integers, and:
x > y
x^2 - 1 > y^4 - 4y + x - 1

Which of the following represents all the possible values of y?

ANSWER:
Let us first simplify the inequality. Cancel out the (-1) from both sides and move the over to the left hand side:
Then factor out the from the left hand side and the on the right hand side:
When we divided both parts of the equation by , we assumed that it is greater than 0. Since at the end we are left without a variable, this means that the equation works for any value where

x(x-1) > y(y-4)
(y+1)(y+1-1) > y(y-4)
y+1 > y-4
1 > 4

Now, any of the integers greater than are part of the solution to this inequality.

The correct answer is B.
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Re: Factoring...? (m02 #8) [#permalink] New post 02 Mar 2008, 14:20
Anyone? :-D
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Re: Factoring...? (m02 #8) [#permalink] New post 02 Mar 2008, 15:27
what is B??

just write the question the way it is statet with the answers. and you will get an answer. :lol:
Re: Factoring...? (m02 #8)   [#permalink] 02 Mar 2008, 15:27
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