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(2) \(2x+5y-20=0\) --> \(2x=20-5y\) --> \(-20+5y>3y\) --> \(y>10\). Same as above: \(x<0\). Sufficient.
Can you please explain stmt. 2 again. Unable to understand the following stmt---
(2) \(2x+5y-20=0\) --> \(2x=20-5y\) --> given \(-2x>3y\), substitute \(2x\) --> \(-(20-5y)>3y\) --> \(-20+5y>3y\) --> \(y>10\) --> \(y=positive\), as discussed above if \(y\) is any positive number then \(x\) must be some negative number: \(x<0\). Sufficient.
Re: If -2x > 3y, is x negative? (1) y > 0 (2) 2x + 5y - 20 = 0 [#permalink]
29 Jun 2013, 06:45
This post received KUDOS
In statement 2 we can write the equation 2x+3y+2y = 20 we know 2x+3y is positive and we get y = 10 hence same as statement 1 is this approach correct?
If -2x > 3y, is x negative?
(1) y > 0 -2x > +ve number, hence x is negative. Sufficient
(2) 2x + 5y - 20 = 0 The area defined by -2x > 3y is the area under the red line. If we know that \(2x + 5y - 20 = 0\) (blue line) (given the initial condition) we can say that x is negative because they intersect when x is negative. (refer to the image) Sufficient
Your approach is correct. We know that 2x+3y is negative (typo I think), so \(2x + 3y +2y= 20\) can be seen as \(-ve +2y=20\) so y is positive for sure as \(2y=20+(+ve)\)
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Re: If -2x > 3y, is x negative? [#permalink]
09 Apr 2015, 13:57
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