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Graphic approach to problems with inequalities

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priyankurml

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11 Mar 2009, 11:12

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walker

Thanks, there is a typo here:
Instead of

walker

... We can put x=4.5 into x-y=2 and find that y=4.25<4.5 (left side).

should be: ... We can put x=4.5 into x-y=2 and find that y=2.5>2.25 (left side, line x-y=2 goes above P).
or even better: ... 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.

I've fixed it in original post.
+1

Hello walker, could you please help me get the final answer? I am clear with the diagram but do not understand how to deduce the answer from it.
intersection of two straight lines is (4.5,2.25).
From first option: x-y<2, I am trying to confirm that whether intersection (4.5,2.25) validates this eqiation.
=> x-y<2
=> y > x-2
=> y > 4.5 - 2
=> y > 2.5
But at x=4.5, actual value of y is 2.25, which is, indeed, not greater than 2.5!! What should I look for from this statement?

I guess I am missing something.....

....thank you.

matrix777

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18 Mar 2009, 20:05

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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:

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13 May 2009, 23:02

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Hi Walker,

This approach works for linear equation or can we use it for equations like X^2 or X^3 also ??

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18 May 2009, 11:24

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mdfrahim

Hi Walker,

This approach works for linear equation or can we use it for equations like X^2 or X^3 also ??

This approach works for any equations that could be drawn under GMAT time conditions. Y = X^3 + 1 is easy to draw but Y = 0.34X^3 - 342X^2 + 6X - 3 is not.

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02 Aug 2009, 04:53

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thanks... but i think it is difficult to use ... ven ur pressed for time

walker

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02 Aug 2009, 10:32

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For me this approach has one great advantage over other ones: When picture is drawn, the answer is almost obvious. Moreover, drawing is pretty straightforward process: line by line, expression by expression.
Nevertheless, the more approaches you know, the more confident you are.

understudy

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10 Aug 2009, 17:43

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Hi Walker - for those of us who are artistically challenged, do you happen to know the way to solve this algebraically?

walker

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10 Aug 2009, 23:03

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understudy

Hi Walker - for those of us who are artistically challenged, do you happen to know the way to solve this algebraically?

I will try to solve it using "construct example" approach:

x/y >2, 3x+2y <18?

a) Could we construct an example when 3x+2y<18 ? We need small x and y for witch x/y >2.
Let's say we have x = 1 (I mean 1.000001) and y=0.5

1/0.5 > 2 - Ok
3*1+2*0.5 <18 Ok

first statement: 1-0.5<2 Ok
second statement: -0.5+1<2 Ok

So, x=1.000001 and y=0.5 is an example that satisfies both statements and answer is YES (True) that 3x + 2y < 18

b) Could we construct an example when 3x+2y>=18 ? The condition x/y > 2 says that x and y can be both positive or negative. At negative x,y 3x+2y will be always negative. So, we need look for our example among x,y positive.

first statement: x-y<2. Under this statement we can say that positive x and y must be close to each other. So, let's consider the maximum difference: x-y = 2 ---> y+2/y >2 --> y=2 and x=4 (as earlier I mean x=3.9999999 y=1.99999999).
At such x and y we have 3*4+ 2*2 = 16 < 18. So, we can't construct an example for witch 3x+2y >=18 and first statement is sufficient.

second statement: y-x < 2. Under this statement we can choose x=100000 and y =1 and 3x+2y >=18. So, we can construct an example for witch 3x+2y >=18 and second statement is insufficient.

My comments: This approach depends on concrete problem and your luck to see an examples. At the same time I love graph approach as it is always straightforward.

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12 Nov 2009, 22:59

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I actually don't know other quick ways to solve it yet...and I am an engineering/math major! I am very fascinated by your approach - most of the time i try to solve problems algebraically.

I got 49 on Quant on my last GMAT exam 2 years ago but sucked on the verbal section- retaking the test on Dec. 14th. Wish me luck. So far my averages on Manhattan CATs, Preps and Peterson are about 670 (highest 700, lowest 650). Quite disappointing but I am still trying to break the hurdle. Working full time is definitely not helping. I even deactivated facebook and turned into a complete anti-social being for the past few months :cry:

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12 Nov 2009, 23:14

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babyif19

Define your weaknesses and spend all your time on them. It is simple but many tend to do what they like but not what they need. If you have problem in SC, thy Manhattan SC book - it is a bible for sentence correction. But anyway, good luck with exam, you are very close to 700.

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30 Dec 2009, 11:23

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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.

akshey2021

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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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28 Oct 2010, 04:24

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akshey2021

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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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

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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graph.png [ 12.14 KiB | Viewed 7040 times ]

utkarshlavania

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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]

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

geno5

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16 Jan 2012, 14:49

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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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26 Oct 2012, 00:37

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matrix777

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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26 Oct 2012, 05:22

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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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11 Mar 2013, 00:16

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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/

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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29 Aug 2013, 21:57

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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.