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

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Re: Is xy > 0? (1) x - y > -2 (2) x - 2y < -6  [#permalink]

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New post 02 Jun 2018, 12:31
AOD wrote:
Bunuel wrote:


We got y > 4 not y = 4, so you cannot substitute the way you did. In the solution, x > 2, is derived by adding y > 4 and x > y - 2 (we can add inequalities when their signs are in the same direction): \(y+x>4+(y-2)\) --> \(x>2\).


Hi Bunuel ,

Kindly clarify -

"In the Solution of X>2" : Instead of adding Y>4 and X > Y-2 (from equation 1) , can we not add Y>4 and -X> 6-2y ? (from equation 2 : x-2y <-6)

If we do this , we will arrive at X < 2


y > 4
-x > 6 - 2y

Adding those gives:
y - x > 10 - 2y
3y - x > 10
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Re: Is xy > 0? (1) x - y > -2 (2) x - 2y < -6  [#permalink]

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New post 02 Jun 2018, 17:42
Bunuel wrote:
AOD wrote:
Bunuel wrote:


We got y > 4 not y = 4, so you cannot substitute the way you did. In the solution, x > 2, is derived by adding y > 4 and x > y - 2 (we can add inequalities when their signs are in the same direction): \(y+x>4+(y-2)\) --> \(x>2\).


Hi Bunuel ,

Kindly clarify -

"In the Solution of X>2" : Instead of adding Y>4 and X > Y-2 (from equation 1) , can we not add Y>4 and -X> 6-2y ? (from equation 2 : x-2y <-6)

If we do this , we will arrive at X < 2


y > 4
-x > 6 - 2y

Adding those gives:
y - x > 10 - 2y
3y - x > 10


Sorry Bunuel , I still don't follow. Maybe I'm missing something.

Last step =3y - x > 10

Substituting Y >4 : 12 - X > 10 -> 2 > X -> X < 2 : right?
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Re: Is xy > 0? (1) x - y > -2 (2) x - 2y < -6  [#permalink]

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New post 03 Jun 2018, 05:10
AOD wrote:
Bunuel wrote:
AOD wrote:

Hi Bunuel ,

Kindly clarify -

"In the Solution of X>2" : Instead of adding Y>4 and X > Y-2 (from equation 1) , can we not add Y>4 and -X> 6-2y ? (from equation 2 : x-2y <-6)

If we do this , we will arrive at X < 2


y > 4
-x > 6 - 2y

Adding those gives:
y - x > 10 - 2y
3y - x > 10


Sorry Bunuel , I still don't follow. Maybe I'm missing something.

Last step =3y - x > 10

Substituting Y >4 : 12 - X > 10 -> 2 > X -> X < 2 : right?


How are you getting 12 - x > 10?

3y - x > 10
3y > 12.

We cannot subtract those two the way you did.

You can only apply subtraction when their signs are in the opposite directions:

If \(a>b\) and \(c<d\) (signs in opposite direction: \(>\) and \(<\)) --> \(a-c>b-d\) (take the sign of the inequality you subtract from).
Example: \(3<4\) and \(5>1\) --> \(3-5<4-1\).

You can only add inequalities when their signs are in the same direction:

If \(a>b\) and \(c>d\) (signs in same direction: \(>\) and \(>\)) --> \(a+c>b+d\).
Example: \(3<4\) and \(2<5\) --> \(3+2<4+5\).

For more check Manipulating Inequalities.
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Re: Is xy > 0? (1) x - y > -2 (2) x - 2y < -6  [#permalink]

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New post 17 Jun 2018, 12:16
Hi,

I tried solving this problem by using substitution method. The problem just says is xy>0, and does not specify if X and Y are Integers or fractions. Following is my solution:-

1) x-y>-2
let x = 4 and y = -2; x-y = 4+2 = 6 >-2
Let x = -1 and y = -.5; x-y = -1 + .5 = .5>-2
Let x = -1 and y = .5; x-y = -1-.5= -1.5>-2
In sufficient

2) x-2y<-6
let x = -2 and y =3; -2-6 = -8<-6
let x = -11 and y = -2; -11 + 4 = -7<-6
let x = 2 and y = 10; 2-20=-18<-6;
In sufficient

1 and 2
If x = -ve, y = -ve; if x = -ve, y = +ve; hence, not sufficient

please tell me where i am wrong
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Re: Is xy > 0 ?  [#permalink]

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New post 04 Oct 2018, 18:45
1
This question is fantastic for bringing up a strategy not enough people think to do with inequalities: as long as you have the inequality signs pointed in the same direction, you can add the inequalities together to eliminate a variable the same way you would for a system of equations.

Here once you've eliminated A, B, and D by quick conceptual number picking**, you can multiply the Statement 2 inequality by -1 so that you have both inequalities with signs in the same direction:

x - y > -2

and

2y - x > 6

If you then add the two inequalities, you eliminate the x term (one is positive and the other is negative), leaving:

y > 4

So you know y is positive, and then when you plug y > 4 in to the first equation (which can be expressed as x > y - 2), you know that x is greater than "something greater than 4, minus 2" so x must also be positive, so you know that the product xy is greater than 0. Together the statements are sufficient and the answer is (C).

Note that when you do have two inequalities this situation presents itself more often than most people think! If you can get the signs facing in the same direction (generally by multiplying one inequality by -1 if they're not already in the same direction) you can add the inequalities together (don't subtract, since "minus" is basically "plus negative" and can screw up the positive/negative inequality logic) and solve like a system of equations.


** Try x = 4 and y = 6 versus x = -2 and y = 0 for statement 1 and x= 0 and y = 3 and x = 1 and y = 12 for statement 2, for example.
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Re: Is xy > 0 ?  [#permalink]

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New post 04 Oct 2018, 19:32
Is xy > 0 ?

This basically asks us .."Are x and y of SAME sign?"

(1) x – y > –2
x>y-2..
y=2, x=4...yes
y=-2, x=4...no
Insufficient

(2) x – 2y < –6
x<2y-6..
y=2, x=-4...no
y=-2, x=-15...yes
Insufficient


Combined
x-y>-2
x-2y<-6 or 2y-x>6
Add the two inequalities
x-y+2y-x>-2+6.......y>4
Multiply first by 2 and add the equations to get x
2(x-y)+2y-x>2*(-2)+6.....2x-2y+2y-x>6-4.....x>2

So both x and y are positive
Ans is always yes
Sufficient

C
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Re: Is xy > 0 ? &nbs [#permalink] 04 Oct 2018, 19:32

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