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Re: Is xy > 0? (1) x - y > -2 (2) x - 2y < -6 [#permalink]
26 May 2012, 02:09

7

This post received KUDOS

Expert's post

3

This post was BOOKMARKED

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.

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)

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 _________________

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. _________________

And many strokes, though with a little axe, hew down and fell the hardest-timbered oak. - William Shakespeare

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. _________________

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.

Answer: C.

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 ?

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.

Answer: C.

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.

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