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x2 > y2 and x > –|y|

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sandy
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x2 > y2 and x > –|y| [#permalink] New post   02 Sep 2018, 07:32
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\(x^2 > y^2\) and \(x > -|y|\)

Quantity A
Quantity B
x
y


A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
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A
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msk0657
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Re: x2 > y2 and x > –|y| [#permalink] New post   03 Sep 2018, 10:26
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sandy wrote:
\(x^2 > y^2\) and \(x > -|y|\)

Quantity A
Quantity B
x
y


A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.


Given \(x^2 > y^2\) and \(x > -|y|\)

From \(x^2 > y^2\) => \(x^2\) - \(y^2\) > 0 => ( x + y ) ( x - y ) > 0.

Here ( x + y ) ( x - y ) > 0 either both should be +ve or both -ve to be greater than 0.

But from \(x > -|y|\) we get ( x + y ) > 0 ( Note here |y| is converted to y )

Since we came to know that ( x + y ) > 0 then must be ( x - y ) > 0

So ( x - y ) > 0 => x > y.

A.
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Yuan
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Re: x2 > y2 and x > –|y| [#permalink] New post   03 Sep 2018, 15:40
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if \(x^2>y^2\), then \(|x|>|y|\)

if x>0 y>0, then x>y since \(|x|>|y|\);

if x>0 y<0, then x>y. Because \(x>-|y|\) and \(-|y|=-(-y)=y\);

x<0 y>0 is not possible. Because \(x>-|y|\) and \(-|y|=-y\), so \(x>-y\), which means \(|x|<|y|\), but we already know \(|x|>|y|\), so this is a contradiction;

x<0 y<0 is also not possbile, Because \(x>-|y|\) and \(-|y|=-(-y)=y\) therefore x>y but then \(|x|<|y|\), and we already know \(|x|>|y|\), so this is a contradiction;
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Re: x2 > y2 and x > –|y| [#permalink] New post   02 Jan 2019, 11:21
Yuan wrote:
if \(x^2>y^2\), then \(|x|>|y|\)

if x>0 y>0, then x>y since \(|x|>|y|\);

if x>0 y<0, then x>y. Because \(x>-|y|\) and \(-|y|=-(-y)=y\);

x<0 y>0 is not possible. Because \(x>-|y|\) and \(-|y|=-y\), so \(x>-y\), which means \(|x|<|y|\), but we already know \(|x|>|y|\), so this is a contradiction;

x<0 y<0 is also not possbile, Because \(x>-|y|\) and \(-|y|=-(-y)=y\) therefore x>y but then \(|x|<|y|\), and we already know \(|x|>|y|\), so this is a contradiction;


\(x>-y\), which means \(|x|<|y|\)
How please explain?
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Re: x2 > y2 and x > –|y| [#permalink] New post   02 Jan 2019, 19:14
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sandy wrote:
\(x^2 > y^2\) and \(x > -|y|\)

Quantity A
Quantity B
x
y


A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.


\(x^2>y^2\) means \(|x|>|y|\)..
This means x is farther away from 0 as compared to y..
So there can be four ways...
x....y....0
X.............0....y
0....y....x
y....0.............x

So if x>0 then x>y , but if x<0, then x<y...
So the relationship depend on the SIGN of x..

Now next we are given x>-|y|
Let us check the four cases
x....y....0. => If x<0 then x>-|y| is not true as -|y| >0.
X..........0....y => If x<0 then x>-|y| is not true as -|y| <0, but -|y| will be closer to 0.
0....y....x => If x>0 then x>-|y| is possible.
y....0..........x => If x<0 then x>-|y| is again possible..

In both the cases that are possible, x>y..

Hence A
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Re: x2 > y2 and x > –|y| [#permalink] New post   25 Jun 2021, 02:00
msk0657 wrote:
sandy wrote:
\(x^2 > y^2\) and \(x > -|y|\)

Quantity A
Quantity B
x
y


A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.


Given \(x^2 > y^2\) and \(x > -|y|\)

From \(x^2 > y^2\) => \(x^2\) - \(y^2\) > 0 => ( x + y ) ( x - y ) > 0.

Here ( x + y ) ( x - y ) > 0 either both should be +ve or both -ve to be greater than 0.

But from \(x > -|y|\) we get ( x + y ) > 0 ( Note here |y| is converted to y )

Since we came to know that ( x + y ) > 0 then must be ( x - y ) > 0

So ( x - y ) > 0 => x > y.

A.


How does: x>−|y| gives -> ( x + y ) > 0
Can someone explain?
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x2 > y2 and x > –|y| [#permalink] New post   25 Jun 2021, 02:30
1
We have take y +ve

\(x > -y\)

\(x + y > 0\)

Check this :
Quote:
x>-|y|
Let us check the four cases
x....y....0. => If x<0 then x>-|y| is not true as -|y| >0.
X..........0....y => If x<0 then x>-|y| is not true as -|y| <0, but -|y| will be closer to 0.
0....y....x => If x>0 then x>-|y| is possible.
y....0..........x => If x<0 then x>-|y| is again possible..

In both the cases that are possible, x>y



botuser wrote:
How does: x>−|y| gives -> ( x + y ) > 0
Can someone explain?
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Re: x2 > y2 and x > –|y| [#permalink] New post   27 Jun 2021, 04:07
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We have been given that the square of x is greater than the square of y. So it must be that the absolute value of x should be greater irrespective of the sign. Now for the sign we can look at the second inequality. here x is greater than the negative of the absolute of y. So it implies that x should be positive because if x is negative and it has the greater absolute value then it should be less than the negative of the absolute value of y.
So this implies that the x is positive and greater than y.
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Re: x2 > y2 and x > |y| [#permalink] New post   02 Sep 2024, 02:22
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Re: x2 > y2 and x > |y| [#permalink]
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