Last visit was: 19 Nov 2025, 08:01 It is currently 19 Nov 2025, 08:01
Home
GMAT
Messages
Tests
Schools
Events
Rewards
Chat
My Profile
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
   1   2   3   4   
User avatar
sher1978
User avatar
Joined: 23 Sep 2009
Last visit: 13 Sep 2010
Posts: 7
Own Kudos:
3
Location: uzbekistan
Concentration: Private Equity
Schools:Chicago booth,Dartmouth,Duke
Posts: 7
Kudos: 3
Graphic approach to problems with inequalities 30 Dec 2009, 11:23
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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.
User avatar
walker
User avatar
Joined: 17 Nov 2007
Last visit: 25 May 2025
Posts: 2,398
Own Kudos:
10,717
 [1]
Given Kudos: 362
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Expert
Expert reply
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Posts: 2,398
Kudos: 10,717
 [1]
Graphic approach to problems with inequalities 31 Dec 2009, 04:54
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
sher1978

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.
Show more
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)
avatar
akshey2021
avatar
Joined: 03 Sep 2010
Last visit: 31 Jul 2011
Posts: 3
Own Kudos:
1
Posts: 3
Kudos: 1
Graphic approach to problems with inequalities 27 Oct 2010, 23:24
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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. ?
User avatar
walker
User avatar
Joined: 17 Nov 2007
Last visit: 25 May 2025
Posts: 2,398
Own Kudos:
10,717
Given Kudos: 362
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Expert
Expert reply
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Posts: 2,398
Kudos: 10,717
Graphic approach to problems with inequalities 28 Oct 2010, 04:24
Kudos
Add Kudos
Bookmarks
Bookmark this Post
akshey2021
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. ?
Show more

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.
User avatar
fivedaysleft
User avatar
Joined: 29 Jun 2011
Last visit: 27 Jun 2014
Posts: 50
Own Kudos:
28
Given Kudos: 46
Posts: 50
Kudos: 28
Graphic approach to problems with inequalities 02 Jul 2011, 04:55
Kudos
Add Kudos
Bookmarks
Bookmark this Post
walker
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

Attachment:
The attachment tarek99.png is no longer available
Show more

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...?
Attachments

graph.png
graph.png [ 12.14 KiB | Viewed 6694 times ]

User avatar
walker
User avatar
Joined: 17 Nov 2007
Last visit: 25 May 2025
Posts: 2,398
Own Kudos:
10,717
 [1]
Given Kudos: 362
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Expert
Expert reply
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Posts: 2,398
Kudos: 10,717
 [1]
Graphic approach to problems with inequalities 02 Jul 2011, 06:28
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
fivedaysleft
i have a doubt...while plotting x/y>2...i multiplied both sides with y and got the inequality x>2y .....
Show more

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.
User avatar
utkarshlavania
User avatar
Joined: 03 Sep 2010
Last visit: 03 Apr 2024
Posts: 10
Own Kudos:
19
Given Kudos: 106
Posts: 10
Kudos: 19
Graphic approach to problems with inequalities 29 Jul 2011, 00:16
Kudos
Add Kudos
Bookmarks
Bookmark this Post
walker


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]
Show more

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
User avatar
geno5
User avatar
Joined: 05 Dec 2011
Last visit: 29 Mar 2022
Posts: 66
Own Kudos:
99
Given Kudos: 2
Location: Canada
Concentration: Accounting, Finance
GMAT Date: 09-08-2012
GPA: 3
Products:
My Reviews
Posts: 66
Kudos: 99
Graphic approach to problems with inequalities 16 Jan 2012, 14:49
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I do not get how to draw (x/y)>2

can someone simplify this? I understand the rest what am I missing?
User avatar
walker
User avatar
Joined: 17 Nov 2007
Last visit: 25 May 2025
Posts: 2,398
Own Kudos:
10,717
 [1]
Given Kudos: 362
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Expert
Expert reply
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Posts: 2,398
Kudos: 10,717
 [1]
Graphic approach to problems with inequalities 16 Jan 2012, 19:58
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
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
avatar
ss58146
avatar
Joined: 25 Oct 2012
Last visit: 11 Nov 2012
Posts: 5
Own Kudos:
1
Given Kudos: 2
Concentration: Economics, Finance
Posts: 5
Kudos: 1
Graphic approach to problems with inequalities 26 Oct 2012, 00:37
Kudos
Add Kudos
Bookmarks
Bookmark this Post
matrix777
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: ?
Show more


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.
User avatar
walker
User avatar
Joined: 17 Nov 2007
Last visit: 25 May 2025
Posts: 2,398
Own Kudos:
10,717
Given Kudos: 362
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Expert
Expert reply
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Posts: 2,398
Kudos: 10,717
Graphic approach to problems with inequalities 26 Oct 2012, 05:22
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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).
User avatar
KarishmaB
User avatar
Joined: 16 Oct 2010
Last visit: 18 Nov 2025
Posts: 16,267
Own Kudos:
76,994
 [2]
Given Kudos: 482
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,267
Kudos: 76,994
 [2]
Graphic approach to problems with inequalities Updated on: 11 Oct 2022, 02:21
Originally posted by KarishmaB on 10 Mar 2013, 20:28.
Last edited by KarishmaB on 11 Oct 2022, 02:21, edited 1 time in total.
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
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.
User avatar
Dipankar6435
User avatar
Joined: 27 Jan 2013
Last visit: 21 Dec 2017
Posts: 58
Own Kudos:
125
Given Kudos: 38
Status:Breaking my head!!
Location: India
Concentration: General Management, Operations
Schools: ISB '16 (A) Ross '17 (D)
GMAT 1: 740 Q50 V40
GPA: 3.51
WE:Other (Transportation)
Schools: ISB '16 (A) Ross '17 (D)
GMAT 1: 740 Q50 V40
Posts: 58
Kudos: 125
Graphic approach to problems with inequalities 11 Mar 2013, 00:16
Kudos
Add Kudos
Bookmarks
Bookmark this Post
VeritasPrepKarishma
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: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/01 ... -part-iii/
Show more

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??
User avatar
KarishmaB
User avatar
Joined: 16 Oct 2010
Last visit: 18 Nov 2025
Posts: 16,267
Own Kudos:
76,994
 [2]
Given Kudos: 482
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,267
Kudos: 76,994
 [2]
Graphic approach to problems with inequalities Updated on: 23 Aug 2023, 00:12
Originally posted by KarishmaB on 11 Mar 2013, 00:49.
Last edited by KarishmaB on 23 Aug 2023, 00:12, edited 1 time in total.
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Dipankar6435

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??
Show more

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.

Check out this post: https://anaprep.com/algebra-inequalitie ... in-action/
and this video: https://youtu.be/VnEVS8kmWa8
User avatar
ygdrasil24
User avatar
Joined: 07 Apr 2012
Last visit: 15 Jan 2014
Posts: 70
Own Kudos:
29
Given Kudos: 45
Location: United States
Concentration: Entrepreneurship, Operations
Schools: ISB '15
GMAT 1: 590 Q48 V23
GPA: 3.9
WE:Operations (Manufacturing)
Schools: ISB '15
GMAT 1: 590 Q48 V23
Posts: 70
Kudos: 29
Graphic approach to problems with inequalities 29 Aug 2013, 21:57
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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.
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,389
Own Kudos:
778,260
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,389
Kudos: 778,260
Graphic approach to problems with inequalities 30 Aug 2013, 04:54
Kudos
Add Kudos
Bookmarks
Bookmark this Post
ygdrasil24
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.
Show more

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.
User avatar
ygdrasil24
User avatar
Joined: 07 Apr 2012
Last visit: 15 Jan 2014
Posts: 70
Own Kudos:
29
Given Kudos: 45
Location: United States
Concentration: Entrepreneurship, Operations
Schools: ISB '15
GMAT 1: 590 Q48 V23
GPA: 3.9
WE:Operations (Manufacturing)
Schools: ISB '15
GMAT 1: 590 Q48 V23
Posts: 70
Kudos: 29
Graphic approach to problems with inequalities 13 Sep 2013, 11:45
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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.
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,389
Own Kudos:
778,260
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,389
Kudos: 778,260
Graphic approach to problems with inequalities 15 Sep 2013, 09:30
Kudos
Add Kudos
Bookmarks
Bookmark this Post
ygdrasil24
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.
Show more

y - x < 2 --> y<x+2. So, it's a whole region below the graph y=x+2 (not limited).
User avatar
ygdrasil24
User avatar
Joined: 07 Apr 2012
Last visit: 15 Jan 2014
Posts: 70
Own Kudos:
29
Given Kudos: 45
Location: United States
Concentration: Entrepreneurship, Operations
Schools: ISB '15
GMAT 1: 590 Q48 V23
GPA: 3.9
WE:Operations (Manufacturing)
Schools: ISB '15
GMAT 1: 590 Q48 V23
Posts: 70
Kudos: 29
Graphic approach to problems with inequalities 15 Sep 2013, 10:27
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
ygdrasil24
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).
Show more
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.
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,389
Own Kudos:
778,260
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,389
Kudos: 778,260
Graphic approach to problems with inequalities 15 Sep 2013, 10:48
Kudos
Add Kudos
Bookmarks
Bookmark this Post
ygdrasil24
Bunuel
ygdrasil24
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.
Show more

It's not a linear equation. So, the approach is different:
walker
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.
Show more
Attachment:
MSP12871igh5183c7965eh7000022622h2c27df8334.gif
MSP12871igh5183c7965eh7000022622h2c27df8334.gif [ 2.71 KiB | Viewed 6337 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
   1   2   3   4   
Moderator:
Bunuel
Math Expert
105389 posts