If -2x>3y, is x negative?

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by Zipper » Thu Dec 04, 2008 4:36 am
1. Obvious, x must be negative.

2. 2x+5y-20=0; 2x+5y=20 | :5;

2/5x+y=4; y=4-2/5x

When we supstitude y in -2x>3y we get -2x>3(4-2/5x); -2x>12-6/5x; -2x+6/5x>12; -0,8x>12 so x must be negative

hence, D

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by austin » Thu Dec 04, 2008 7:30 am
If -2x>3y, is x negative?

Stem translation: if x < (-3/2)y, is x <0??


(1) y>0 => x is negative....

(2) 2x+5y-20=0
2x + 5y = 20
x < -1.5 y.

put x = -2y => y = 20; x<-30=> x is negative...
you can try for different values...

D

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by cramya » Thu Dec 04, 2008 4:10 pm
Another way(without picking numbers)t


Question stem

-2x>3y => 0>3y+2x (i.e 3y+2x is negative)

Stmt I

y>0

Then x has to be negative so that the inequality -2x>3y holds

SUFF

Stmt II

2x+5y-20=0
2x+5y = 20

2x+3y+2y=20

Since we know 2x+3y<0 therefore 2y has to be positive since the result is 20 (positive)

So we again know y>0 therefore x<0 again

SUFF


Hope I dint mess up with stmt II anlaysis :-)

D)

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by lunarpower » Fri Dec 05, 2008 3:18 am
here's a more ground-level solution for statement (2).

notice that there's a '2x' in the inequality given in the question prompt, and another '2x' in statement 2. this is a fine opportunity for substitution.

so, SOLVE both things for the '2x':

divide the inequality by -1 --> 2x < -3y

isolate the '2x' in statement (2) --> 2x = -5y + 20

substitute --> -5y + 20 < -3y

solve --> 20 < 2y
10 < y
y > 10
which means
-2x > 3y > 30
-2x > 30
x < -15
sufficient

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you can theoretically use plug-in-numbers on this problem, but it's much more trouble than is worthwhile.
Ron has been teaching various standardized tests for 20 years.

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by 4meonly » Fri Dec 05, 2008 4:05 am
Ron, thanx.
I dont like to plug numbers. This is second-best approach for me.

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by farooq » Thu Oct 29, 2009 5:04 am
you can theoretically use plug-in-numbers on this problem, but it's much more trouble than is worthwhile.
As you said and shown, once we simply the inequality we can plug-in-numbers comfortably.

Thanks for this valuable explanation.
Regards,
Farooq Farooqui.
London. UK

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