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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 \(5y > 0\).

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 (>).

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.

Re: If x and y are integers and x > 0, is y > 0? [#permalink]
15 Oct 2013, 01:29

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Re: If x and y are integers and x > 0, is y > 0? [#permalink]
03 May 2014, 20:22

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.

Re: If x and y are integers and x > 0, is y > 0? [#permalink]
04 May 2014, 02:09

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.

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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