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Math Expert V
Joined: 02 Sep 2009
Posts: 58464
Re: Is xy > 0? (1) x - y > -2 (2) x - 2y < -6  [#permalink]

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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

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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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

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?
Math Expert V
Joined: 02 Sep 2009
Posts: 58464
Re: Is xy > 0? (1) x - y > -2 (2) x - 2y < -6  [#permalink]

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1
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

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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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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Joined: 19 Jul 2018
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Re: Is xy > 0 ?  [#permalink]

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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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Math Expert V
Joined: 02 Aug 2009
Posts: 8023
Re: Is xy > 0 ?  [#permalink]

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1
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? (1) x - y > -2 (2) x - 2y < -6  [#permalink]

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Baten80 wrote:
Is xy > 0?

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

Target question: Is xy > 0?

Statement 1: x - y > -2
Let's TEST some values.
There are several values of x and y that satisfy statement 1. Here are two:
Case a: x = 5 and y = 1. In this case, xy = (5)(1) = 5. So, the answer to the target question is YES, xy IS greater than 0
Case b: x = 5 and y = -1. In this case, xy = (5)(-1) = -5. So, the answer to the target question is NO, xy is NOT greater than 0
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: x - 2y < -6
Let's TEST some values again.
There are several values of x and y that satisfy statement 2. Here are two:
Case a: x = 1 and y = 10. In this case, xy = (1)(10) = 10. So, the answer to the target question is YES, xy IS greater than 0
Case b: x = -1 and y = 10. In this case, xy = (-1)(10) = -10. So, the answer to the target question is NO, xy is NOT greater than 0
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
IMPORTANT: There’s a useful inequality property that says If two inequalities have their inequality symbols facing the SAME DIRECTION, we can ADD the inequalities.
We have:
x - y > -2
x - 2y < -6

Take the BOTTOM equation and multiply both sides by -1 to get:
x - y > -2
-x + 2y > 6 [since we multiplied both sides by a NEGATIVE value, we REVERSED the direction of the inequality symbol]

Now ADD the equations to get: y > 4
In other words, y is POSITIVE

Now take:
x - y > -2
-x + 2y > 6

Take the TOP equation and multiply both sides by 2 to get:
2x - 2y > -4
-x + 2y > 6

Now ADD the equations to get: x > 2
In other words, x is POSITIVE

Since x and y are both POSITIVE, we know that xy IS positive
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

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

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Baten80 wrote:
Is xy > 0?

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

Is xy>0

1) x-y > -2
x>y-2
xy>y(y-2)=y^2-2y
Value of y is unknown
NOT SUFFICIENT

2) x-2y < -6
x<2y-6
xy<y(2y-6)=2y^2-6y
Value of y is unknown
NOT SUFFICIENT

Combining (1) & (2)
1) x-y > -2
x>y-2. (3)
2) x-2y < -6
x<2y-6
-x>-2y+6 (4)
Adding (3) & (4)
0>-y+4
y>4>0 (5)
From equation (3)
x>y-2>4-2=2>0
x>2>0 (6)
xy>0
SUFFICIENT

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

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