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

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Orange08
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If x and y are integers and x > 0, is y > 0? [#permalink] New post  Updated on: 20 Oct 2019, 00:31
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If x and y are integers and x > 0, is y > 0?

(1) 7x – 2y > 0
(2) -y < x


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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?
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Originally posted by Orange08 on 11 Sep 2010, 15:05.
Last edited by Bunuel on 20 Oct 2019, 00:31, edited 2 times in total.
Renamed the topic and edited the question.
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Bunuel
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Re: If x and y are integers and x > 0, is y > 0? [#permalink] New post  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.
Two different answers. Not sufficient.

Answer: E.

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: If x and y are integers and x > 0, is y > 0? [#permalink] New post  21 Aug 2011, 13:49
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samramandy wrote:
If x and y are integers and x > 0, is y > 0?
(1) 7x-2y > 0
(2) -y < x


1. 7x > 2y
x=2; y<7 will hold true for y=1, y=0, y=-1
Not Sufficient.

2. x+y>0
x=2; y>-2 will hold true for y=-1, y=0, y=1
Not Sufficient.

Combining both:
7x-2y+x+y>0
8x-y>0
8x>y
x=2; y<16; y can be 1,0,-1
Not Sufficient.

Ans: "E"
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Orange08
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Re: If x and y are integers and x > 0, is y > 0? [#permalink] New post  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: If x and y are integers and x > 0, is y > 0? [#permalink] New post  11 Sep 2010, 15:34
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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?


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

Answer C.
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Re: If x and y are integers and x > 0, is y > 0? [#permalink] New post  04 May 2014, 03:09
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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.

Answer C.


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] New post  11 Feb 2017, 01:17
1) 7x-2y>0

7(8)-2(3) = positive
7(8)-2(-3) = positive

y can be neg or pos. NOT sufficient.

2) -y<x

x = 8 , y can equal 3
x = 8, -3<8, so y can be positive

x = 8, y can equal -3
x=8, -(-3)<8, 6<8, y can be negative too. Not sufficient.

c) combined: Look at the breakdown from statement (1) with numbers that also satisfy statement (2)'s criteria. y can be neg OR positive.

Answer is Choice E.
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Re: If x and y are integers and x > 0, is y > 0? [#permalink] New post  27 Jul 2017, 06:11
from 1 alone
y can be negative and y can be positive
say y=-1 and x=1.
7+2=9. 9>0

say y=1 and x=1
7-2=5. 5>0

1 alone is insufficient. A and D are out.

statement 2 says -y<x.
rewrite 0<x+y
y can be negative and y can be positive.
say y=-1 and x=2
2-1 = 1. 1>0. works

say y=1 and x=1
1+1=2. 2>0. works.

B is out, so we are left with either C or E - 50/50 chance we'd get it right, not bad
but combining, we see that both statements say that y can be > or < 0.
C is out, and E is the answer.
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Re: If x and y are integers and x > 0, is y > 0? [#permalink] New post  26 Sep 2018, 18:49
y=-1 ,x =1

kills 1, 2 and both
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Re: If x and y are integers and x > 0, is y > 0? [#permalink] New post  14 May 2023, 16:26
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