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Is x² - y > 1/y² where y is non-zero?

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Re: Is x² - y > 1/y² where y is non-zero? [#permalink]
PriyankaPalit7 wrote:
Is x² - y > 1/y² where y is non-zero?

(1) x < y
(2) x² < y²

$$x^2$$ - y > 1/$$y^2$$

$$x^2$$ - y - 1/$$y^2$$ > 0

Lets try 2 first
(2) x² < y² or y^2 > x^2,
9 > 4, y can be -3 or 3 and x can be -2 or 2
1/4 > 1/16, y can be 1/2 or -1/2 and x can be 1/4 or -1/4

4 +3 - 1/9 > 0, This answers a yes to the question
1/16 - 1/2 - 1/4 < 0, This answers a no to the question

from 1)
x < y or y>x, here y & x can be again a fraction, integer or a number
1/2 > 1/4 or 3 > 2
1/16 - 1/2 - 1/4 < 0, This answers a no to the question
4 -3 - 1/9 > 0, This answers a yes to the question

The same test cases can be used again to show that after combining the 2, we wont get an unique answer

E
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Is x² - y > 1/y² where y is non-zero? [#permalink]
PriyankaPalit7 wrote:
Is x² - y > 1/y² where y is non-zero?

(1) x < y
(2) x² < y²

Hi arpitkansal,

What $$x^2<y^2$$ means is $$|x|<|y|$$..
X<y and |x|<|y| when combined mean y>x and y>0..
By looking at the equation, $$x^2-y>\frac{1}{y^2}$$, we should realise that $$\frac{1}{y^2}$$ will be positive.
We have to see if x^2-y can be both negative and positive.
1) x=2 and y=3
So $$2^2-3>\frac{1}{3^2}....4-3>\frac{1}{9}...1>\frac{1}{9}....$$yes
2) x=1 and y=4
$$1^2-4>\frac{1}{4^2}.....1-4>\frac{1}{16}.....-3>\frac{1}{16}...$$.no

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Re: Is x² - y > 1/y² where y is non-zero? [#permalink]
PriyankaPalit7 wrote:
Is x² - y > 1/y² where y is non-zero?

(1) x < y
(2) x² < y²

Take (x,y) = (0,1) & (0,-1).

We get E.
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Re: Is x² - y > 1/y² where y is non-zero? [#permalink]
rahul16singh28 wrote:
PriyankaPalit7 wrote:
Is x² - y > 1/y² where y is non-zero?

(1) x < y
(2) x² < y²

Take (x,y) = (0,1) & (0,-1).

We get E.

Hi Rahul, be careful with numbers you pick up..
(o, -1) does not fit into statement 1, which is x<y.. as 0>-1, meaning x>y
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Is x² - y > 1/y² where y is non-zero? [#permalink]
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Here's how I solved, feel free to critique!

Is $$x^{2}$$ - y > $$\frac{1}{y^{2}}$$? If $$y\neq{0}$$.

I simplified the expression as follows:

Because $$y^{2}$$ is always positive I multiplied both sides by $$y^{2}$$ to get $$x^{2}y^{2}$$ - $$y^{3}$$ > 1.

From here I factored out $$y^{2}$$ to get $$y^{2}$$($$x^{2}$$ - y) > 1.

The question is now:

Is $$y^{2}$$($$x^{2}$$ - y) > 1? If Y $$x\neq{0}$$.

Statement 1: x < y

I picked x as -1 and y as 1. Plugging these values into the question the result is 0 > 1.

I then picked x as -3 and y as -2. Plugging these values into the question the result is 44 > 1.

Clearly insufficient.

Statement 2: $$x^{3}$$ < $$y^{3}$$

I picked x as -1 and y as 1. Since they are raised to an odd power the signs remain the same for x and y (1 and -1, respectively).
Plugging these values into the question the result is 0 > 1.

I then picked x as -3 and y as -2. Since they are raised to an odd power the signs remain the same for x and y (-3 and -2, respectively).
Plugging these values into the question the result is 44 > 1.

Clearly insufficient.

Statement 1&2

Clearly insufficient.

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Re: Is x² - y > 1/y² where y is non-zero? [#permalink]
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PriyankaPalit7 wrote:
Is x² - y > 1/y² where y is non-zero?

(1) x < y
(2) x² < y²

(1) x < y Let x = -3, y =-2. Then we have (-3)^2 --2 > 1/(-2)^2 --> is 9 -2 > 1/4? Yes

Let x =-1/2 let y =-1/4 is (-1/2)^2 --1/4 > 1/(-1/4)^2 ? is 1/4 - 1/4 >4 NO NS

(2) x^2 <y^2 This is equivalent to |x| <|y| Let x =-2 y =-3

Is (-2)^2 --3 > 1/(-3)^2 ? is 4+3 > 1/9? Yes

let x = 3 y =4 is (3)^2 -4 > 1/4^2? is 9-4 > 1/16 No

NS

(1) and (2) x < y and |x|<|y| so values like x = -4, y =-3 won't work Let x=0 y =1 is 0^2 -1 >1/1^2 No

Let x =3, y=4 is 3^2 -4 > 1/4^2 ? is 9-4 >1/4^2 Yes

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Is x² - y > 1/y² where y is non-zero? [#permalink]
PriyankaPalit7 wrote:
Is x² - y > 1/y² where y is non-zero?

(1) x < y
(2) x² < y²

Hey VeritasKarishma Bunuel, could this be solved algebraically?

$$x^2-y>1/y^2…x^2y^2-y^3>1?$$
$$[I.]…x=2,y=3:(4)(9)-(27)>1?…36-27>1?…answer=yes$$
$$[II.]…x=1,y=4:(1)(16)-(64)>1?…16-64>1?…answer=no$$

(1) x < y: case [I.] and [II.] are possible; insufic.
(2) x² < y²: case [I.] and [II.] are possible; insufic.
(1&2): case [I.] and [II.] are possible; insufic.

Originally posted by exc4libur on 08 Oct 2019, 08:22.
Last edited by exc4libur on 09 Oct 2019, 07:20, edited 1 time in total.
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Re: Is x² - y > 1/y² where y is non-zero? [#permalink]
exc4libur wrote:
PriyankaPalit7 wrote:
Is x² - y > 1/y² where y is non-zero?

(1) x < y
(2) x² < y²

Hey VeritasKarishma Bunuel, could this be solved algebraically?

$$x^2-y>1/y^2…x^2y^2-y^3>1$$
$$[I.]…x=1,y=2:(1)(4)-8>1?…4-8=-4<1…answer=no$$
$$[II.]…x=-3,y=-4:(9)(16)-(64)>1?…144-64=80>1…answer=yes$$

(1) x < y: case [I.] and [II.] are possible; insufic.
(2) x² < y²: case [I.] and [II.] are possible; insufic.
(1&2): case [I.] and [II.] are possible; insufic.

You are still using number plugging (which is pretty much the way to go in this question).
Also, how come the two examples work for both statements? When x = -3 and y = -4, x > y, not x < y

x < y and x^2 < y^2 when x and y are both positive.
Try
x = 1, y = 2
x = 1/4, y = 1/2
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Re: Is x² - y > 1/y² where y is non-zero? [#permalink]

You are still using number plugging (which is pretty much the way to go in this question).
Also, how come the two examples work for both statements? When x = -3 and y = -4, x > y, not x < y

x < y and x^2 < y^2 when x and y are both positive.
Try
x = 1, y = 2
x = 1/4, y = 1/2

There I fixed it. Thanks
But can you solve this algebraically? Or is it a waste of time?
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Re: Is x² - y > 1/y² where y is non-zero? [#permalink]
exc4libur wrote:

You are still using number plugging (which is pretty much the way to go in this question).
Also, how come the two examples work for both statements? When x = -3 and y = -4, x > y, not x < y

x < y and x^2 < y^2 when x and y are both positive.
Try
x = 1, y = 2
x = 1/4, y = 1/2

There I fixed it. Thanks
But can you solve this algebraically? Or is it a waste of time?

I wouldn't even try algebra here.
x < y and x^2 < y^2 should remind you of number properties. For what kind of numbers does this hold? For non negative numbers. We know that numbers behave differently 'between 0 and 1' and '1 onwards'. So I will try both type of numbers.
Re: Is x² - y > 1/y² where y is non-zero? [#permalink]
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