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Given: -2x>3y. Q: is x<0? (Note here that if y is any positive number then we would have -2x>positive, and in order that to be true x must be some negative number).
(1) y>0 --> -2x>3y>0 --> x<0. Sufficient.
(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
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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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