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.
basic OG question
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- dblazquez
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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?
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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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.
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.
- dblazquez
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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
Daniel
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
but the question is askingTherefore, no points that satisfy the inequiality are in quadrant IV
so I think that there is an error...is that an OG question? 11th or 10th? what do you guys think?which quadrant system satisfy the inequity
Daniel
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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
![Smile :)](./images/smilies/smile.png)
Yes, this is from 10th OG and I scanned the OG and attached the file.
Thanks for your help!
Jane
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- dblazquez
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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?![Sad :(](./images/smilies/sad.png)
![Smile :)](./images/smilies/smile.png)
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?
![Sad :(](./images/smilies/sad.png)
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hi, i am a new comer. i am trying my best to learn Gmat but skills are so baddblazquez wrote:By the way, the area that satisfies the inequation is the area above the graph...
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
- sk818020
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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.
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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ohh i get it. thanks alotsk818020 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.
![Smile :)](./images/smilies/smile.png)
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.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.
Thanks alot