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basic OG question

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by magical cook » Wed Aug 02, 2006 1:13 am
sorry for the very basic question but I dont understand the explanation for the answer.....

Thank you.
Jane

which quadrant system satisfy the inequity 2X - 3Y < -6

1. I
2. II
3. III
4. IV
5. None

since when Y is negative, X is also neative. (till here I understand..) Therefore, the no point (x,y) can have negative Y and positive X. (why???) The answer is 4 IV.
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by dblazquez » Thu Aug 03, 2006 12:29 am
I've tried the picking numbers strategy:

Cuad/x/y/value
I 2 2 -2
II -2 2 -10 <<
III 2 -2 10
IV -2 -2 2

so IMO i'd say (2) II, but i am not sure at all, for picking 1's yields a different result.... what is the OA? explanations?
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explanation from OG

by magical cook » Thu Aug 03, 2006 12:45 am
Thank you for your time.... here is explanation from OG.


The answer is 4. the explanation on OG says..."Since 2X-3Y<-6, 2X<3Y, or X<3Y-6/2, note taht if Y is negative, X must be negative. Therefore, no point (x,y) that satisfies the inequality can have Y negative and X positive. Therefore, no points that satisfy the inequiality are in quadrant IV. The best answer is 4.
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by dblazquez » Thu Aug 03, 2006 2:09 am
thnx to you for posting:)

Now i dont understand neither... I think that something is wrong there... probably the question stem is not accurate. According to the explanation, it seems that we are looking for points that don't satisfy the inequation
Therefore, no points that satisfy the inequiality are in quadrant IV
but the question is asking
which quadrant system satisfy the inequity
so I think that there is an error...is that an OG question? 11th or 10th? what do you guys think?

Daniel
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by magical cook » Thu Aug 03, 2006 2:49 am
Thanks for your time too.:)

Yes, this is from 10th OG and I scanned the OG and attached the file.

Thanks for your help!
Jane
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by dblazquez » Thu Aug 03, 2006 3:12 am
okay there is subtle difference between "which cuadrant contains no point that satisfies" and "which quadrant system satisfy the inequity" but makes it almost opposite :) perhaps that difference is why you're confused...

anyway, it's a tricky one... I have plotted the graph to see how is the line. According to the graph, only the IV makes it work, but i dont know how to figure out without the graph...maybe picking numbers? :(
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by dblazquez » Thu Aug 03, 2006 3:14 am
By the way, the area that satisfies the inequation is the area above the graph...
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by diebeatsthegmat » Fri May 07, 2010 9:37 am
dblazquez wrote:By the way, the area that satisfies the inequation is the area above the graph...
hi, i am a new comer. i am trying my best to learn Gmat but skills are so bad
anyways, for this question, i dont really understand something.
agree that if y is negative, x must be , however what would happen if y can be positive?
for an exam. 2x-3y<-6
x=1 y=6, thus 2-18=-16<-6 thus it could be in the first I
x=-7 y= -2, thus -14+6=-8<-6, so it could be in the III
now i am really confused
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by sk818020 » Fri May 07, 2010 1:03 pm
This one might be a little easier to solve if you put the inequality into y=mx+b format.

2X - 3Y < -6
-3Y<-2x-6
Y>(2/3)x+2

Solving for when x=0 and y=0. You get (0,4/3), (-3,0). Put those to point on a graph. Drawa line connecting them. Y is >, so all the points above the line satisfy the equation. Clearly none in the fourth quadrant satisfy the equation.

This is probably the best way to solve it. You risk messing up your calculations and taking more time when your testing in numbers on a question like this.
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by diebeatsthegmat » Mon May 10, 2010 9:06 am
sk818020 wrote:This one might be a little easier to solve if you put the inequality into y=mx+b format.

2X - 3Y < -6
-3Y<-2x-6
Y>(2/3)x+2

Solving for when x=0 and y=0. You get (0,4/3), (-3,0). Put those to point on a graph. Drawa line connecting them. Y is >, so all the points above the line satisfy the equation. Clearly none in the fourth quadrant satisfy the equation.

This is probably the best way to solve it. You risk messing up your calculations and taking more time when your testing in numbers on a question like this.
ohh i get it. thanks alot :)
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by bearcat » Tue May 11, 2010 8:16 am
sk818020 wrote:This one might be a little easier to solve if you put the inequality into y=mx+b format.

2X - 3Y < -6
-3Y<-2x-6
Y>(2/3)x+2

Solving for when x=0 and y=0. You get (0,4/3), (-3,0). Put those to point on a graph. Drawa line connecting them. Y is >, so all the points above the line satisfy the equation. Clearly none in the fourth quadrant satisfy the equation.

This is probably the best way to solve it. You risk messing up your calculations and taking more time when your testing in numbers on a question like this.
I appreciate this approach. Could you clarify more on your explanation that as Y>, all the point above the line satisfy the equation. A systematic explaination, in my opinion, offer lots help in the sames questions.

Thanks alot
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