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If x and y are integers and x > 0, is y > 0?

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If x and y are integers and x > 0, is y > 0? [#permalink]

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11 Sep 2010, 15:05
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If x and y are integers and x > 0, is y > 0?

(1) 7x – 2y > 0
(2) -y < x

From 1, y can have +ve and -ve values. For example, x = 2, y = 1 or x = 2, y = -1. Thus, not sufficient to answer whether y >0

From 2, -y < x or x +y > 0, again not sufficient. x = 0.9, y = -0.8 or x = 0.9, y = 0.8

Combining 1 & 2
7x - 2y > 0
x + y > 0

----------------

7x - 2y > 0
7x + 7y > 0 (multiplying earlier equation 2 by 7)

summing the above two equations, can't we derive 5y > 0 and thus y > 0?
[Reveal] Spoiler: OA
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Re: Is y > 0? [#permalink]

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11 Sep 2010, 15:17
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Orange08 wrote:
If x and y are integers and x > 0, is y > 0?
(1) 7x – 2y > 0
(2) -y < x

From 1, y can have +ve and -ve values. For example, x = 2, y = 1 or x = 2, y = -1. Thus, not sufficient to answer whether y >0

From 2, -y < x or x +y > 0, again not sufficient. x = 0.9, y = -0.8 or x = 0.9, y = 0.8

Combining 1 & 2
7x - 2y > 0
x + y > 0

----------------

7x - 2y > 0
7x + 7y > 0 (multiplying earlier equation 2 by 7)

summing the above two equations, can't we derive 5y > 0 and thus y > 0?

If $$y=0$$ and $$x$$ is any positive integer both statements hold true and the answer to the question is NO.
If $$y=1$$ and $$x=2$$ again both statements hold true and the answer to the question is YES.

As for your question: when we sum $$7x - 2y > 0$$ and $$7x + 7y > 0$$ we'll get $$14x+5y>0$$ and not $$5y > 0$$.

Hope it helps.
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Re: Is y > 0? [#permalink]

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11 Sep 2010, 15:26
Bunuel wrote:
Orange08 wrote:
x + y > 0

----------------

As for your question: when we sum $$7x - 2y > 0$$ and $$7x + 7y > 0$$ we'll get $$14x+5y>0$$ and not $$5y > 0$$.

Hope it helps.

Thanks. Oh ya,, silly mistake from me. I meant subtracting equation 2 from 1 will lead us to
9y >0 and eventually y > 0.

Am i wrong in performing such subtractions for inequalities?
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Re: Is y > 0? [#permalink]

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11 Sep 2010, 15:34
Orange08 wrote:
Bunuel wrote:
Orange08 wrote:
x + y > 0

----------------

As for your question: when we sum $$7x - 2y > 0$$ and $$7x + 7y > 0$$ we'll get $$14x+5y>0$$ and not $$5y > 0$$.

Hope it helps.

Thanks. Oh ya,, silly mistake from me. I meant subtracting equation 2 from 1 will lead us to
9y >0 and eventually y > 0.

Am i wrong in performing such subtractions for inequalities?

I don't think simultaneous equations are solved on inequalities.

If A > 0 and B> 0; it doesn't mean A-B>0. A=2, B 5 will prove this.
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Re: Is y > 0? [#permalink]

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11 Sep 2010, 15:39
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Orange08 wrote:
Bunuel wrote:
Orange08 wrote:
x + y > 0

----------------

As for your question: when we sum $$7x - 2y > 0$$ and $$7x + 7y > 0$$ we'll get $$14x+5y>0$$ and not $$5y > 0$$.

Hope it helps.

Thanks. Oh ya,, silly mistake from me. I meant subtracting equation 2 from 1 will lead us to
9y >0 and eventually y > 0.

Am i wrong in performing such subtractions for inequalities?

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 $$2<5$$ --> $$3+2<4+5$$.

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 $$5>1$$ --> $$3-5<4-1$$.

So we can only add $$7x - 2y > 0$$ and $$7x + 7y > 0$$ as their signs are in the same direction (>).

Hope it helps.
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Re: Is y > 0? [#permalink]

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11 Sep 2010, 16:57
Nice. I didn't think this for inequalities.
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Re: Is y > 0? [#permalink]

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12 Sep 2010, 01:37
Bunuel wrote:

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 $$2<5$$ --> $$3+2<4+5$$.

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 $$5>1$$ --> $$3-5<4-1$$.

So we can only add $$7x - 2y > 0$$ and $$7x + 7y > 0$$ as their signs are in the same direction (>).

Hope it helps.

Absolutely fantastic. Thanks Bunuel once again. This has indeed cleared one of the flaws in my understanding.
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09 Feb 2012, 00:42
According to my doc, the OA is E.
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Re: If x and y are integers and x > 0; is y > 0 ? (1) 7x - [#permalink]

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09 Feb 2012, 01:12
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Re: If x and y are integers and x > 0, is y > 0? [#permalink]

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15 Oct 2013, 02:29
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Re: If x and y are integers and x > 0, is y > 0? [#permalink]

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03 May 2014, 21:22
If x and y are integers and x > 0, is y > 0?

(1) 7x – 2y > 0
(2) -y < x

Bunuel, please have a look at this.

statement1. 7x - 2y > 0
=> x > 2y/7---------(1)

y can be negative or positive. Not sufficient

statement2. -y < x
=> x + y > 0---------(2)

Again, y can be negative or positive. Not Sufficient.

statement 1 & 2 Together.

from eq. (1) & (2),
[a value > 2y/7] + y > 0
lets consider least value of x (actually this would be lesser than the least).
2y/7 + y > 0
=> 9y/7 > 0 or (9/7)y>0
therefore, y must be > than 0.

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Re: If x and y are integers and x > 0, is y > 0? [#permalink]

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04 May 2014, 03:09
Expert's post
honey86 wrote:
If x and y are integers and x > 0, is y > 0?

(1) 7x – 2y > 0
(2) -y < x

Bunuel, please have a look at this.

statement1. 7x - 2y > 0
=> x > 2y/7---------(1)

y can be negative or positive. Not sufficient

statement2. -y < x
=> x + y > 0---------(2)

Again, y can be negative or positive. Not Sufficient.

statement 1 & 2 Together.

from eq. (1) & (2),
[a value > 2y/7] + y > 0
lets consider least value of x (actually this would be lesser than the least).
2y/7 + y > 0
=> 9y/7 > 0 or (9/7)y>0
therefore, y must be > than 0.

Why are you considering the least value of x??? Nothing prevents x to be greater than this value and in this case your logic will not hold.
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Re: If x and y are integers and x > 0, is y > 0? [#permalink]

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31 Oct 2015, 12:12
Hello from the GMAT Club BumpBot!

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Re: If x and y are integers and x > 0, is y > 0?   [#permalink] 31 Oct 2015, 12:12
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