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Given: \(-2x>3y\). Question: 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).

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

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).

(2) \(2x+5y-20=0\) --> \(2x=20-5y\) --> \(-20+5y>3y\) --> \(y>10\). Same as above: \(x<0\). Sufficient.

Answer: D.

Can you please explain stmt. 2 again. Unable to understand the following stmt---

\(-20+5y>3y\)

(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]

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29 Jun 2013, 06:45

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This post received KUDOS

fozzzy wrote:

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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If -2x > 3y, is x negative? (1) y > 0 (2) 2x + 5y - 20 = 0

-2x > 3y 2x + 3y<0 -----(1)

Statement 1 If y>0 & 2x + 3y<0

Then x must be Negative. Sufficient

Statement 2 2x + 5y - 20 = 0 2x + 5y = 20 (2x + 3y) + 2y=20 We can write 2y + some negative no = 20 2y = 20 + some Positiveno y = 10 + some Positiveno/2 This mean that y>10

2x + 3y<0 2x< -3y x < -1.5 (Positive no) because y is positive

Then x must be Negative. Sufficient

Answer D
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If -2x > 3y, is x negative? (1) y > 0 (2) 2x + 5y - 20 = 0

Hi, -2x > 3y... (a)If y<0, x can be both +ive and -ive.. (b)if y>0, x will have to be +ive as 3y is positive and -2x , to be positive, has to have x as -ive..

now lets see the choices.. (1) y > 0 If y>0, x is -ive as proved in (b) above... suff

(2) 2x + 5y - 20 = 0.. this can be written as 2x+3y + 2y -20=0.. now 2x+3y<0, so 2y>20... or y is +ive and therefore x is -ive.... suff

If -2x > 3y, is x negative? (1) y > 0 (2) 2x + 5y - 20 = 0

Hi, -2x > 3y... (a)If y<0, x can be both +ive and -ive.. (b)if y>0, x will have to be +ive as 3y is positive and -2x , to be positive, has to have x as -ive..

now lets see the choices.. (1) y > 0 If y>0, x is -ive as proved in (b) above... suff

(2) 2x + 5y - 20 = 0.. this can be written as 2x+3y + 2y -20=0.. now 2x+3y<0, so 2y>20... or y is +ive and therefore x is -ive.... suff

If -2x > 3y, is x negative? (1) y > 0 (2) 2x + 5y - 20 = 0

Hi, -2x > 3y... (a)If y<0, x can be both +ive and -ive.. (b)if y>0, x will have to be +ive as 3y is positive and -2x , to be positive, has to have x as -ive..

now lets see the choices.. (1) y > 0 If y>0, x is -ive as proved in (b) above... suff

(2) 2x + 5y - 20 = 0.. this can be written as 2x+3y + 2y -20=0.. now 2x+3y<0, so 2y>20... or y is +ive and therefore x is -ive.... suff

ans D

how can you say 2x+3y<0?

Hi, 2x + 3y <0 comes from -2x>3y.. -2x>3y.. add 2x to both sides.. 2x-2x>3y+2x.. 0>2x+3y... hope it helps
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Hi, 2x + 3y <0 comes from -2x>3y.. -2x>3y.. add 2x to both sides.. 2x-2x>3y+2x.. 0>2x+3y... hope it helps

Aye, it does! Thanks

Also, if I understand, <> sign changes in multiplication only![/quote]

hi, yes you are right , whenever you multiply two sides on either side of equality with a -ive sign or -ive quantity, you are required to change the greater/lesser than sign.. -2x>3y.. 2x<-3y..
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