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If x and y are integers and x > 0, is y > 0?
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Updated on: 27 Jul 2017, 06:07
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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?
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Originally posted by Orange08 on 11 Sep 2010, 15:05.
Last edited by mvictor on 27 Jul 2017, 06:07, edited 1 time in total.
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Re: If x and y are integers and x > 0, is y > 0?
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11 Sep 2010, 15:17
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?
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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: If x and y are integers and x > 0, is y > 0?
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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 AB>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?
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11 Sep 2010, 15:39
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 \(<\)) > \(ac>bd\) (take the sign of the inequality you subtract from). Example: \(3<4\) and \(5>1\) > \(35<41\). 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?
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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 \(<\)) > \(ac>bd\) (take the sign of the inequality you subtract from). Example: \(3<4\) and \(5>1\) > \(35<41\).
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?
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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.
Answer C.



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Re: If x and y are integers and x > 0, is y > 0?
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04 May 2014, 03:09



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Re: If x and y are integers and x > 0, is y > 0?
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11 Feb 2017, 01:17
1) 7x2y>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?
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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 72=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 21 = 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?
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26 Sep 2018, 18:49
y=1 ,x =1
kills 1, 2 and both




Re: If x and y are integers and x > 0, is y > 0? &nbs
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26 Sep 2018, 18:49






