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Graphic approach to problems with inequalities

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Re: Graphic approach to problems with inequalities  [#permalink]

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New post 30 Dec 2009, 10:23
HI there!


Walker, thank you for yuor post.Can you clarify some points,as i could not get the point:

How to identify true or false regions?

Why condition 1 is sufficient and condition 2 is not.As to me both of them have some points on true region and some on false region.
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New post 31 Dec 2009, 03:54
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sher1978 wrote:
How to identify true or false regions?

Why condition 1 is sufficient and condition 2 is not.As to me both of them have some points on true region and some on false region.

Let's say you have y>2x+1. You draw line y=2x+1 and above region is TRUE, below is FALSE. If you have some doubts about that, you may check any point from regions. For example, point (1,100) is above and 100>2+1 is TRUE.

No, only second condition has points on true and false regions. All points for first condition are on true region. (see figures)
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Re: Graphic approach to problems with inequalities  [#permalink]

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New post 27 Oct 2010, 22:24
Apologies for asking such a basic question - I'm afraid i haven't yet figured out how to draw these graphs accurately in the actual exam- I've heard you aren't permitted to take a ruler into the center.

How do you draw the figure for eg. x-y < 2 accurately enough to see that the equation passes through point P in your diag. ?
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Re: Graphic approach to problems with inequalities  [#permalink]

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New post 28 Oct 2010, 03:24
akshey2021 wrote:
Apologies for asking such a basic question - I'm afraid i haven't yet figured out how to draw these graphs accurately in the actual exam- I've heard you aren't permitted to take a ruler into the center.

How do you draw the figure for eg. x-y < 2 accurately enough to see that the equation passes through point P in your diag. ?


Yes, you can't use a ruler. In many cases even if you don't draw accurate, you will see key points and areas. Then (like in 4. here) you can calculate for a couple of questionable points their precise positions.
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Re: Graphic approach to problems with inequalities  [#permalink]

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New post 02 Jul 2011, 03:55
walker wrote:
Hi all! My friend, Tarek, PM me and asked me to show how to use the graphic approach to problem with inequalities. I really love such approach because it is not only fast one after training, but also reliable. So, I try to illustrate how to use it.

1) If \((x/y)>2\), is \(3x+2y<18?\)

(1) \(x-y\) is less than \(2\)
(2) \(y-x\) is less than \(2\)

1. First of all, we draw x/y>2. x/y=2 - is a boundary. (see figure 1). we should note that if one of the variables is negative and other is positive, x/y will be always negative and less than 2. Therefore, our set of x,y that satisfied x/y>2 lies between line x/y=2 and x-axis.

2. Next, we draw our main inequality: 3x+2y<18. 3x+2y=18 - is a boundary. (see figure 2).

3. Now, we should combine our main inequality with the restriction, x/y>2. (see figure 3). Eventually, we defined two areas (sets) were the main inequality is TRUE and were it is FALSE. Two lines intersect in point P with coordinates: (4.5;2.25).

4. Let's consider fist condition: x-y<2. x-y=2 is a boundary. (see figure 4). As we can see all y,x that satisfies the fist condition lie in "green-TRUE" region. Therefore, the first statement is sufficient to answer the question. We should be careful and check where line x-y=2 passes point P, through left side or right side. We can put y=2.25 into x-y=2 and find that x=4.25<4.5 (left side). In other words, line x-y=2 passes y=2.25 (y-coordinate of P) early and goes above P.

5. Finally, let's check last condition: y-x<2. y-x=2 is a boundary. (see figure 5). As we can see all y,x that satisfies the second condition lie in both "green-TRUE" and "red-FALSE" regions. Thus, the second condition is insufficient.

So, answer is A

This approach took less than 2 minutes.

Tips:

1) How fast can we draw a line, for example 3x+2y=18? Simple approach: we need two points to draw line, let's choose intersections with x- and y- axes. x=0 (intersection with y-axis) --> y=9; y=0 (intersection with x-axis) --> x=6.

2) Let's suppose we have a linear inequality, such as 38y-11x>121, suppose we've already drawn the line. How can we find what side is "true" and what side is "false"? The fastest method is just use y=0,x=-infinity. In our case, 0-(-infinity)=infinity>121 - true. Therefore, we take a left side.


That's all :)
Regards,
Serg a.k.a. Walker

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The attachment tarek99.png is no longer available



i have a doubt...while plotting x/y>2...i multiplied both sides with y and got the inequality x>2y and followed nach0's rules... i got this graph...what am i missing here? whats wrong...?
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New post 02 Jul 2011, 05:28
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fivedaysleft wrote:
i have a doubt...while plotting x/y>2...i multiplied both sides with y and got the inequality x>2y .....


Your problem is a typical one for inequalities and modulus questions. You just forget to consider the case when y is negative.
Here is what you should do:

1. x/y >2
2. x >2y (y>0) & x <2y (y<0)
3. solve both inequalities BUT don't forget to apply conditions (y>0 and y<0). For example, in your plot y can't be negative.

By the way, try to check out whether the answer makes sense. Moreover, sometimes it's useful think a bit about expression at the beginning. For example, x/y > 2 only if x and y have the same sign. So, your last graph fails to pass this test.

Actually, that is why I used graphic approach as it allows to avoid such kind of mistakes.
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New post 28 Jul 2011, 23:16
walker wrote:

1. First of all, we draw x/y>2. x/y=2 - is a boundary. (see figure 1). we should note that if one of the variables is negative and other is positive, x/y will be always negative and less than 2. Therefore, our set of x,y that satisfied x/y>2 lies between line x/y=2 and x-axis.

[/url]


Hello Walker and everyone
I might sound really dumb here but just wanted to clarify a doubt as to WHY NOT any point in quadrant 1 and quadrant 3 relating to (X,Y) will give x/y>2. How did you calculate that the points "x/y>2 lies between line x/y=2 and x-axis."
I understood that both need to be negative or both the variables need to be Positive. But these values seem to be satisfied by points in whole of quadrant 1 and quadrant 3

Waiting for your reply
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Re: Graphic approach to problems with inequalities  [#permalink]

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New post 16 Jan 2012, 13:49
I do not get how to draw (x/y)>2

can someone simplify this? I understand the rest what am I missing?
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New post 16 Jan 2012, 18:58
x/y >2 is not the same as x > 2y and here is how to approach it in a right way:

There are 3 options:
1) y>0: x > 2y ---> y < 1/2x
2) y=0: undefined
3) y<0: x < 2y ---> y > 1/2x

1) area above y=0 and below y = 1/2x
2) we are not drawing it but keeping in mind that at y=0 the inequality is undefined
3) area below y=0 and above y=1/2x
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New post 25 Oct 2012, 23:37
matrix777 wrote:
Ok this might be a dumb question, but I see both graphs 4 and 5 lines passing from the true region.
How can you deduce which one is correct?

Also for graph 4 "walker" said, "4. Let's consider fist condition: x-y<2. x-y=2 is a boundary. (see figure 4). As we can see all y,x that satisfies the fist condition lie in "green-TRUE" region."

But I see some points that lie outside, how can "all" points satisfy this?

I m in waiting list, gotta take the test again to get in. Please please can somebody help me :shock: ?



same questions here. can you please explain the "all" , that u mentioned walker? since point (10,8) don't satisfy the green region.

Thank you.
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New post 26 Oct 2012, 04:22
Don't forget about x/y>2 that defines area of possible values. (10,8) is not one of them as 10/8 < 2.

Let me sum up the main approach using the set theory:

Our problem:

If A is true, is B true?
1) C is true
2) D is true

Approach

A - defines the set of all possible values. At this point we forget all other values outside of A as they are not considered in the problem
B - divides A by two subsets: A(true) and A(false)
C - we need to figure out whether the intersection of A and C (A ∩ C) has elements only A(true) or only A(false)
D - the same as for C

So, when I said "all" I meant all possible values (it doesn't include any elements outside of A).
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New post 10 Mar 2013, 19:28
Responding to a pm:

To convert x/y > 2 to y = kx form, you can take two cases:

Case 1:
y > 0
x > 2y

Case 2:
y < 0
x < 2y

Now draw x = 2y (see the first diagram of the first post by walker. It's the blue line)
When y > 0 (i.e. region above the x axis), x > 2y. Just plug in a point to figure out the required region. Say, the inequality holds for (4, 1) and hence the required region is below the blue line (but above the x axis)

Similarly, when y < 0 (i.e. region below the x axis), x < 2y. Again, the region between the x axis and x = 2y will be the required region.

So you have got the entire blue region. This is where the relation x/y > 2 holds.

For more on plotting regions of inequality, check out this post: http://www.veritasprep.com/blog/2011/01 ... -part-iii/
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New post 10 Mar 2013, 23:16
VeritasPrepKarishma wrote:
Responding to a pm:

To convert x/y > 2 to y = kx form, you can take two cases:

Case 1:
y > 0
x > 2y

Case 2:
y < 0
x < 2y

Now draw x = 2y (see the first diagram of the first post by walker. It's the blue line)
When y > 0 (i.e. region above the x axis), x > 2y. Just plug in a point to figure out the required region. Say, the inequality holds for (4, 1) and hence the required region is below the blue line (but above the x axis)

Similarly, when y < 0 (i.e. region below the x axis), x < 2y. Again, the region between the x axis and x = 2y will be the required region.

So you have got the entire blue region. This is where the relation x/y > 2
For more on plotting regions of inequality, check out this post: http://www.veritasprep.com/blog/2011/01 ... -part-iii/


Those were helpful indeed :-D. Thanks. Your Quarter Wit, Quarter Wisdom blog is an eye opener. However i feel that question could have been solved faster by an algebric approach, but only because we were provided that both x and y are positive. In absence of such constraints i think the graphical approach will be faster. Another query though. I found this mentioned at the end of the post. I hope you have come to appreciate the wide range of applicability of graphs. Next time, I will introduce a graphical way of working with Modulus and Inequalities. Could I get a link for that. Thnx in advance
And i faintly remember learning some tricks about plotting graphs in my high school. Something to the tune of-"Graph of y=kx can be drawn by expanding the graph of y=x by k times and y=K+x can be drawn by shifting y=x by k units(to the left or right?? :roll: ) Do i need to revisit those for the GMAT??
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Re: Graphic approach to problems with inequalities  [#permalink]

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New post 10 Mar 2013, 23:49
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1
Dipankar6435 wrote:
Those were helpful indeed :-D. Thanks. Your Quarter Wit, Quarter Wisdom blog is an eye opener. However i feel that question could have been solved faster by an algebric approach, but only because we were provided that both x and y are positive. In absence of such constraints i think the graphical approach will be faster. Another query though. I found this mentioned at the end of the post. I hope you have come to appreciate the wide range of applicability of graphs. Next time, I will introduce a graphical way of working with Modulus and Inequalities. Could I get a link for that. Thnx in advance
And i faintly remember learning some tricks about plotting graphs in my high school. Something to the tune of-"Graph of y=kx can be drawn by expanding the graph of y=x by k times and y=K+x can be drawn by shifting y=x by k units(to the left or right?? :roll: ) Do i need to revisit those for the GMAT??


Sure, you can use either method - it depends on what you are more comfortable with. I find working with equations/inequalities way too cumbersome and have developed an ease with graphs (with practice of course). I prefer to take a holistic view and figure out the answer since GMAT questions are basically logic based, (and hence the ample use of graphs). You might find that graphs slow you down initially but with practice, they can save you a lot of time. Anyway, both the methods work perfectly fine so choose whichever you like more.
I have many posts on Mods and inequalities peppered in-between other posts on my blog. I would suggest you to start from the bottom of the last page and go upwards checking out the posts that catch your fancy: http://www.veritasprep.com/blog/categor ... om/page/3/

And yes, those tricks help you approach the questions keeping the big picture in mind. In fact I have discussed some of them here:
http://www.veritasprep.com/blog/2010/12 ... he-graphs/

They are not essential to know if you plan to use algebra for most questions. They can be quite helpful if you plan on working out the questions using the holistic approaches.
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Re: Graphic approach to problems with inequalities  [#permalink]

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New post 29 Aug 2013, 20:57
Attachment:
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First of all, hats off, I have tried myself building such approaches but could never get a full proof version that can cover all possible types, urs do.
I have one doubt,

using (i) , x-y<2 will be a region which will pull down the line or the region underneath it ? , So then false condition will also start coming in that region
the region marked grey will come in < condition
using(ii) y-x<2 will be a region which will pull down the line or the region underneath it ? , So then false condition will also start coming in that region which looks ok to me.

Point out if I am interpreting it wrongly.
Thanks.
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New post 30 Aug 2013, 03:54
ygdrasil24 wrote:
Attachment:
tarek99.png


First of all, hats off, I have tried myself building such approaches but could never get a full proof version that can cover all possible types, urs do.
I have one doubt,

using (i) , x-y<2 will be a region which will pull down the line or the region underneath it ? , So then false condition will also start coming in that region
the region marked grey will come in < condition
using(ii) y-x<2 will be a region which will pull down the line or the region underneath it ? , So then false condition will also start coming in that region which looks ok to me.

Point out if I am interpreting it wrongly.
Thanks.


Rewrite x-y<2 as y>x-2. True region is ABOVE (because of ">" sign) the line y=x-2.
Rewrite y-x<2 as y<x+2. True region is BELOW (because of "<" sign) the line y=x+2.

This question is also discussed here: if-x-y-2-is-3x-2y-18-1-x-y-is-less-than-89225.html

Hope it helps.
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Re: Graphic approach to problems with inequalities  [#permalink]

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New post 13 Sep 2013, 10:45
I have a question, when we say y - x < 2. So we consider region pointing downward as < y condition. But what is the limit of that region.
As per my understanding, its upto infinity, unless crossed out by another line, say x = -5, so our area of consideration gets reduced. Please correct me if I am wrong.
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New post 15 Sep 2013, 08:30
ygdrasil24 wrote:
I have a question, when we say y - x < 2. So we consider region pointing downward as < y condition. But what is the limit of that region.
As per my understanding, its upto infinity, unless crossed out by another line, say x = -5, so our area of consideration gets reduced. Please correct me if I am wrong.
Thanks.


y - x < 2 --> y<x+2. So, it's a whole region below the graph y=x+2 (not limited).
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Re: Graphic approach to problems with inequalities  [#permalink]

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New post 15 Sep 2013, 09:27
Bunuel wrote:
ygdrasil24 wrote:
I have a question, when we say y - x < 2. So we consider region pointing downward as < y condition. But what is the limit of that region.
As per my understanding, its upto infinity, unless crossed out by another line, say x = -5, so our area of consideration gets reduced. Please correct me if I am wrong.
Thanks.


y - x < 2 --> y<x+2. So, it's a whole region below the graph y=x+2 (not limited).

OK. If you refer to the link graphic-approach-to-problems-with-inequalities-68037.html
by Walker, how is the graph x/y >2 drawn here? y < x/2 so graph should be all region less than y = x/2. I didnot get how this graph is drawn.
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Re: Graphic approach to problems with inequalities  [#permalink]

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New post 15 Sep 2013, 09:48
ygdrasil24 wrote:
Bunuel wrote:
ygdrasil24 wrote:
I have a question, when we say y - x < 2. So we consider region pointing downward as < y condition. But what is the limit of that region.
As per my understanding, its upto infinity, unless crossed out by another line, say x = -5, so our area of consideration gets reduced. Please correct me if I am wrong.
Thanks.


y - x < 2 --> y<x+2. So, it's a whole region below the graph y=x+2 (not limited).

OK. If you refer to the link graphic-approach-to-problems-with-inequalities-68037.html
by Walker, how is the graph x/y >2 drawn here? y < x/2 so graph should be all region less than y = x/2. I didnot get how this graph is drawn.


It's not a linear equation. So, the approach is different:
walker wrote:
1. First of all, we draw x/y>2. x/y=2 - is a boundary. (see figure 1). we should note that if one of the variables is negative and other is positive, x/y will be always negative and less than 2. Therefore, our set of x,y that satisfied x/y>2 lies between line x/y=2 and x-axis.
Attachment:
MSP12871igh5183c7965eh7000022622h2c27df8334.gif
MSP12871igh5183c7965eh7000022622h2c27df8334.gif [ 2.71 KiB | Viewed 3219 times ]


If you are not comfortable with this method use algebra: if-x-y-2-is-3x-2y-18-1-x-y-is-less-than-89225.html
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Re: Graphic approach to problems with inequalities   [#permalink] 15 Sep 2013, 09:48

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