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Re: |x|y > x|y| [#permalink]
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Re: |x|y > x|y| [#permalink]
Why can't you square root both sides?
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Re: |x|y > x|y| [#permalink]
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haeshadae wrote:
Why can't you square root both sides?


it is not the way to solve this question.

Please refer to the explanation above by Brent
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Re: |x|y > x|y| [#permalink]
Carcass


Hello Carcass!

I did know that Y was +ve and x -ve but, I my mistake was to apply square root for both sides.

Why is not possible to do that in this question?, cuz we will have + and- values?

Regards!
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Re: |x|y > x|y| [#permalink]
FCOCGALVAN wrote:


Hello Carcass!

I did know that Y was +ve and x -ve but, I my mistake was to apply square root for both sides.

Why is not possible to do that in this question?, cuz we will have + and- values?

Regards!


Take an example which satisfies the given inequality
\(x = -2\) and \(y = 1\)

\(|-2|(1) > (-2)|1|\) = TRUE

Now I suppose you would want to take square roots;

Col. A: \(x + y\)
Col. B: \(x - y\)

Plug in the values;

Col. A: \((-2) + 1 = -1\)
Col. B: \((-2) - 1 = -3\)

Col. A > Col. B

But, actually it's other way around:

Col. A: \((-2 + 1)^2 = (-1)^2 = 1\)
Col. B: \((-2 - 1)^2 = (-3)^2 = 9\)

Col. A < Col. B

That is a direct contradiction!
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Re: |x|y > x|y| [#permalink]
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Thank you Sir
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Re: |x|y > x|y| [#permalink]
FCOCGALVAN wrote:
Carcass


Hello Carcass!

I did know that Y was +ve and x -ve but, I my mistake was to apply square root for both sides.

Why is not possible to do that in this question?, cuz we will have + and- values?

Regards!



if you apply square root, modulus value should be taken care of,

I hope this is your doubt!
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Re: |x|y > x|y| [#permalink]
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Re: |x|y > x|y| [#permalink]
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