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# Is xy > 0? (1) x - y > -2 (2) x - 2y < -6

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Is xy > 0? (1) x - y > -2 (2) x - 2y < -6 [#permalink]  05 Jun 2011, 04:06
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Question Stats:

62% (02:02) correct 38% (01:03) wrong based on 371 sessions
Is xy > 0?

(1) x - y > -2
(2) x - 2y < -6
[Reveal] Spoiler: OA

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Last edited by Bunuel on 09 Jul 2014, 09:49, edited 2 times in total.
Edited the question.
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Re: Is xy > 0? (1) x - y > -2 (2) x - 2y < -6 [#permalink]  05 Jun 2011, 04:20
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(1)

x = 1, y = 0

xy = 0 although x-y = 1 > -2

x = -3, y = -2

xy = 6 > 0 and x-y = -1 > -2

Insufficient

(2)

x - 2y < -6

x = 1, y = 4

x - 2y = 1 - 8 = -7 < -6 and xy > 0

x = 0, y = 4

x - 2y = 0 - 8 = -8 < -6 and xy = 0

Insufficient

(1) + (2)

x - y > -2

2y - x > 6

y > 4

Now x has to be such that x > y - 2

So x is positive, because y is 4.1, 5 etc.
and x is > 2.1, 3 etc.

So xy = positive

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Re: Is xy > 0? (1) x - y > -2 (2) x - 2y < -6 [#permalink]  05 Jun 2011, 23:36
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a+b

x-y > -2 and -x+2y >6 (changing the signs for addition)

y>4 and x >2.1.

Thus xy>0 always. C
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Re: Is xy > 0? (1) x - y > -2 (2) x - 2y < -6 [#permalink]  25 May 2012, 18:15
Hi all,

The process that I went through was

reverse the sign of the second equation

-x + 2y > 6 (to make the signs same for both equations)

and

solve for y to get y>4

Now for solving x

Do you solve the equations again for x

or

how does it work by substituting y > 4 ? Could someone show this method and also tell me if this is the best way to solve these questions.

Thanks
Ravi
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Re: Is xy > 0? (1) x - y > -2 (2) x - 2y < -6 [#permalink]  26 May 2012, 02:09
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Expert's post
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raviram80 wrote:
Hi all,

The process that I went through was

reverse the sign of the second equation

-x + 2y > 6 (to make the signs same for both equations)

and

solve for y to get y>4

Now for solving x

Do you solve the equations again for x

or

how does it work by substituting y > 4 ? Could someone show this method and also tell me if this is the best way to solve these questions.

Thanks
Ravi

Is xy>0?

Note that question basically asks whether $$x$$ and $$y$$ have the same sign.

(1) x-y > -2 --> we can have an YES answer, if for example $$x$$ and $$y$$ are both positive ($$x=10$$ and $$y=1$$) as well as a NO answer, if for example $$x$$ is positive and $$y$$ is negative ($$x=10$$ and $$y=-10$$). Not sufficient.

(2) x-2y <-6 --> again it' easy to get an YES answer, if for example $$x$$ and $$y$$ are both positive ($$x=1$$ and $$y=10$$) as well as a NO answer, if for example $$x$$ is negative and $$y$$ is positive ($$x=-1$$ and $$y=10$$). Not sufficient.

You can get that the the two statement individually are not sufficient in another way too: we have (1) $$y<x+2$$ and (2) $$y>\frac{x}{2}+3$$. We are asked whether $$x$$ and $$y$$ have the same sign or whether the points (x,y) are in the I or III quadrant ONLY. But all (x,y) points below the line $$y=x+2$$ (for 1) and all (x, y) points above the line $$y=\frac{x}{2}+3$$ can not lie only I or III quadrant: points above or below some line (not parallel to axis) lie at least in 3 quadrants.

(1)+(2) Now, remember that we can subtract inequalities with the signs in opposite direction --> subtract (2) from (1): $$x-y-(x-2y)>-2-(-6)$$ --> $$y>4$$. As $$y>4$$ and (from 1) $$x>y-2$$ then $$x>2$$ (because we can add inequalities when their signs are in the same direction, so: $$y+x>4+(y-2)$$ --> $$x>2$$) --> we have that $$y>4$$ and $$x>2$$: both $$x$$ and $$y$$ are positive. Sufficient.

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Re: Is xy > 0? (1) x - y > -2 (2) x - 2y < -6 [#permalink]  12 Mar 2013, 09:28
Bunuel wrote:
tech3 wrote:
Hello, I have a doubt.

After we have got y >4, do we check both the equations for consistency? With the first equation, m getting x >2 but with the second, I am receiving x<2 upon solving. What am I doing wrong

Can you please show how you get x<2?

_____________________________________________________________________________________________________________________________
Bunuel, I'm struggling to understand how the second statement proves that x is definitely positive. I follow that y>4 and I follow that x>2 if I only test equation (1). But I would think that I should be able to test either and come away with the fact that x is positive. However when I test equation (2) x proves to be positive or negative!

Testing equation (2) x - 2y < -6

since y>4
If y = 4.1, then x - 8.2 < -6 and x < 2.2 (this suggests x can be positive or negative)
If y = 6, then x - 12 < -6 and x < 6 (this suggests x can be positive or negative)
and so on...

I don't understand how we prove x is definitely positive if I test equation (2)
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Re: Is xy > 0? (1) x - y > -2 (2) x - 2y < -6 [#permalink]  18 Mar 2013, 12:04
adding both equations, you would get y >4. Substituting this to equation 1, you would get x > 2! Hence both x and y are positive, so xy > 0
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Re: Is xy > 0? (1) x - y > -2 (2) x - 2y < -6 [#permalink]  15 Jul 2013, 16:12
pleaman wrote:
dolumoks wrote:
Bunuel wrote:
Hello, I have a doubt.

After we have got y >4, do we check both the equations for consistency? With the first equation, m getting x >2 but with the second, I am receiving x<2 upon solving. What am I doing wrong

Can you please show how you get x<2?

_____________________________________________________________________________________________________________________________
Bunuel, I'm struggling to understand how the second statement proves that x is definitely positive. I follow that y>4 and I follow that x>2 if I only test equation (1). But I would think that I should be able to test either and come away with the fact that x is positive. However when I test equation (2) x proves to be positive or negative!

Testing equation (2) x - 2y < -6

since y>4
If y = 4.1, then x - 8.2 < -6 and x < 2.2 (this suggests x can be positive or negative)
If y = 6, then x - 12 < -6 and x < 6 (this suggests x can be positive or negative)
and so on...

I don't understand how we prove x is definitely positive if I test equation (2)

Agree! I cant understand the same )

I have the same doubt, with y>4 , put the value in 2nd equation, and you have -negative values for x.
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Re: Is xy > 0? (1) x - y > -2 (2) x - 2y < -6 [#permalink]  15 Jul 2013, 21:37
Expert's post
nikhiil wrote:
pleaman wrote:
dolumoks wrote:
_____________________________________________________________________________________________________________________________
Bunuel, I'm struggling to understand how the second statement proves that x is definitely positive. I follow that y>4 and I follow that x>2 if I only test equation (1). But I would think that I should be able to test either and come away with the fact that x is positive. However when I test equation (2) x proves to be positive or negative!

Testing equation (2) x - 2y < -6

since y>4
If y = 4.1, then x - 8.2 < -6 and x < 2.2 (this suggests x can be positive or negative)
If y = 6, then x - 12 < -6 and x < 6 (this suggests x can be positive or negative)
and so on...

I don't understand how we prove x is definitely positive if I test equation (2)

Agree! I cant understand the same )

I have the same doubt, with y>4 , put the value in 2nd equation, and you have -negative values for x.

But if you use y>4 and x-y > -2 you'll get that x>2. It's more restrictive so it's correct.
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Re: Is xy > 0? (1) x - y > -2 (2) x - 2y < -6 [#permalink]  17 Jul 2013, 10:24
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Please find attached graphical solution. I think its easier and simpler to solve such inequality problem using co-ordinate geometry.

I made this graph in a hurry in paint, so please don't mind the poorly drawn lines.

Consider pressing on Kudos if my post helped you in any way!
Attachments

Solution.JPG [ 39.4 KiB | Viewed 10574 times ]

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Re: Is xy > 0? (1) x - y > -2 (2) x - 2y < -6 [#permalink]  17 Jul 2013, 22:34
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Expert's post

Graphic approach:

Attachment:

xy.png [ 8.35 KiB | Viewed 11475 times ]

Is xy > 0?

$$xy>0$$ means that x and y must have the same sign, so point (x, y) must be either in the first or the third quadrant (green regions).

(1) x-y > -2 --> $$y<x+2$$ --> the area below blue line ($$y=x+2$$). (x, y) may or may not be in green region. Not sufficient.

(2) x-2y < -6 --> $$y>\frac{x}{2}+3$$ --> the area above red line ($$y>\frac{x}{2}+3$$). (x, y) may or may not be in green region. Not sufficient.

(1)+(2) Below blue line and above red line, is yellow region, which is entirely in I quadrant (where $$y>4$$ and $$x>2$$) --> $$xy>0$$. Sufficient.

Hope it helps.
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Re: Is xy > 0? (1) x - y > -2 (2) x - 2y < -6 [#permalink]  31 Jul 2014, 06:19
Bunuel wrote:

Graphic approach:

Attachment:
xy.png

Is xy > 0?

$$xy>0$$ means that x and y must have the same sign, so point (x, y) must be either in the first or the third quadrant (green regions).

(1) x-y > -2 --> $$y<x+2$$ --> the area below blue line ($$y=x+2$$). (x, y) may or may not be in green region. Not sufficient.

(2) x-2y < -6 --> $$y>\frac{x}{2}+3$$ --> the area above red line ($$y>\frac{x}{2}+3$$). (x, y) may or may not be in green region. Not sufficient.

(1)+(2) Below blue line and above red line, is yellow region, which is entirely in I quadrant (where $$y>4$$ and $$x>2$$) --> $$xy>0$$. Sufficient.

Hope it helps.

Hi Bunuel,

I have doubts on graphical method -->

why is line y<x+2 passing through x = -2 ?
the value of x should be greater than -2 (the line may pass through x = -1, 0, 1, 2, ...)
Similarly, why is line y > (x/2) + 3 passing through x = -6
the value of x should be less than -6 (the line may pass through x = -7, -8, -9,...)

With the graphical method, can you illustrate solution of following example ?

Is x + y > 0 ?

1) x - y > 0
2) x^2 - y^2 > 0

Regards,
Akshay
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Re: Is xy > 0? (1) x - y > -2 (2) x - 2y < -6 [#permalink]  31 Jul 2014, 06:40
Expert's post
AkshayDavid wrote:
Bunuel wrote:

Graphic approach:

Is xy > 0?

$$xy>0$$ means that x and y must have the same sign, so point (x, y) must be either in the first or the third quadrant (green regions).

(1) x-y > -2 --> $$y<x+2$$ --> the area below blue line ($$y=x+2$$). (x, y) may or may not be in green region. Not sufficient.

(2) x-2y < -6 --> $$y>\frac{x}{2}+3$$ --> the area above red line ($$y>\frac{x}{2}+3$$). (x, y) may or may not be in green region. Not sufficient.

(1)+(2) Below blue line and above red line, is yellow region, which is entirely in I quadrant (where $$y>4$$ and $$x>2$$) --> $$xy>0$$. Sufficient.

Hope it helps.

Hi Bunuel,

I have doubts on graphical method -->

why is line y<x+2 passing through x = -2 ?
the value of x should be greater than -2 (the line may pass through x = -1, 0, 1, 2, ...)
Similarly, why is line y > (x/2) + 3 passing through x = -6
the value of x should be less than -6 (the line may pass through x = -7, -8, -9,...)

With the graphical method, can you illustrate solution of following example ?

Is x + y > 0 ?

1) x - y > 0
2) x^2 - y^2 > 0

Regards,
Akshay

y < x + 2 is the region which lies BELOW (because of < sign) line y = x + 2.

y > x/2 + 3 is the region which lies ABOVE (because of > sign) line y = x/2 + 3.

Check here for more: graphic-approach-to-problems-with-inequalities-68037.html

Hope it helps.
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Re: Is xy > 0? (1) x - y > -2 (2) x - 2y < -6 [#permalink]  25 Aug 2014, 05:22
Here is an alternate method I used to solve the combined case (data in choice a + data in choice b)

Given
(1) x - y > -2
(2) x - 2y < -6

rewriting

(1) x>y-2 OR y-2<x
(2) x<2y-6

Combining inequalities

y-2<x<2y-6

This equation provides all legal values of x,y. Plugging in numbers, you see that this inequality is true only for positive values of y

Ex: y=-3
-5<x<-12 Not true: -5 cannot be less than -12

Ex: y=0
-2<x<-6 Not true

Only positive values of y work here and since x is greater than y, x also has to be positive
Re: Is xy > 0? (1) x - y > -2 (2) x - 2y < -6   [#permalink] 25 Aug 2014, 05:22
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