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Graphic approach to problems with inequalities

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Graphic approach to problems with inequalities [#permalink]

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29 Jul 2008, 04:51
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Hi all! My friend Tarek PMed me and asked to show how to use the graphic approach to problems with inequalities. I really love the approach because it isn't only fast one after training, but also reliable. So, I try to illustrate how to use it.

There are two distinct ways to solve math problems: logical and visual. The approach described in this post is for those who use the visual approach. If you like imaging math problems in the visual scratchpad in your mind, you will enjoy this method. At the same time, don't be discourage if it doesn't work for you, try the logical approach instead (described by Bunuel).

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.

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,
Walker

Attachment:

tarek99.png [ 17.04 KiB | Viewed 43635 times ]

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

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mainhoon wrote:
Can someone solve this without the graphical approach? Bunuel?

My solution from: tough-inequality-challange-89225.html?hilit=walker%20graphic#p732298

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

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

$$\frac{x}{y}>2$$ tells us that $$x$$ and $$y$$ are either both positive or both negative, which means that all points $$(x,y)$$ satisfying given inequality are either in I or III quadrant. When they are both negative (in III quadrant) inequality $$3x+2y<18$$ is always true, so we should check only for I quadrant, or when both $$x$$ and $$y$$ are positive.

In I quadrant, as $$x$$ and $$y$$ are both positive, we can rewrite $$\frac{x}{y}>2$$ as $$x>2y>0$$ (remember $$x>0$$ and $$y>0$$).

So basically question becomes: If $$x>0$$ and $$y>0$$ and $$x>2y>0$$, is $$3x+2y<18$$?

(1) $$x-y<2$$.

Subtract inequalities $$x>2y$$ and $$x-y<2$$ (we can do this as signs are in opposite direction) --> $$x-(x-y)>2y-2$$ --> $$y<2$$.

Now add inequalities $$x-y<2$$ and $$y<2$$ (we can do this as signs are in the same direction) --> $$x-y+y<2+2$$ --> $$x<4$$.

We got $$y<2$$ and $$x<4$$. If we take maximum values $$x=4$$ and $$y=2$$ and substitute in $$3x+2y<18$$, we'll get $$12+4=16<18$$.

Sufficient.

(2) $$y-x<2$$ and $$x>2y$$:
$$x=3$$ and $$y=1$$ --> $$3x+2y=11<18$$ true.
$$x=11$$ and $$y=5$$ --> $$3x+2y=43<18$$ false.

Not sufficient.

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29 Jul 2008, 05:02
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This is a so powerful approach that I decided to spent some time to illustrate it. Good luck all with your exam!
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Re: Graphic approach to problems with inequalities [#permalink]

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03 Aug 2008, 10:27
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Hi - can someone help me explain Tip #2? I don't really understand what it is saying. Thanks.

See fig.2 - we have line 3x+2y=18. Where is 3x+2y<18? left or right side? we put x=-infinity and y=0 --> -infinity<18. Is is correct? Yes. Therefore, left side corresponds to 3x+2y<18 (orange color in fig.2)
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Re: Graphic approach to problems with inequalities [#permalink]

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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.

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:
tarek99.png

There is another easier method to solve such questions. Let's take the same example to illustrate the method.

given : x > 2y

so we can write x = 2y + k where k is a positive number - equation 1

Fact 1 : x - y < 2

we can rewrite this as x -y + m =2 where m is a positive number - equation 2

solving equation 1 and equation 2

y = 2 -m - k AND x = 2 - m + y = (4 -2m -k)

=> 3x + 2y = 12 - 6m -3k + 4 - 2m - 2k = 16 -8m - 5k

so, 3x + 2y = 16 -8m -5k which is less than 18

now considering fact 2

Fact 2 : y - x < 2

we can rewrite this as y -x +m = 2 - equation 3

solving equation 1 and equation 3

y = m -2 -k and x = 2m -k -4

=> 3x + 2y = 8m -5k -16 which may or may not be less than 18 (since m and k may be of any positive value)

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

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03 Aug 2008, 09:34
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Hi - can someone help me explain Tip #2? I don't really understand what it is saying. Thanks.
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07 Mar 2009, 11:28
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kbulse wrote:
I didn't know that , i was hesitating whether to ask that question, now I am glad that I asked about that. Thanks

No problem that is why we all are here
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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 x=4.5 into x-y=2 and find that y=4.25<4.5 (left side).

Hi ! Can you please clarify the above part for me. According to me, to check for point p, if we put x=4.5 and y = 2.25, we get 4.5-2.25=2.25, which does not satisfy the inequality x-y<2 and thus lie in the false region. I am a little confused here, trying my hands at this for the first time, please correct me if I am wrong.
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Thanks, there is a typo here:
walker wrote:
... 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
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Re: Graphic approach to problems with inequalities [#permalink]

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priyankur_saha@ml.com wrote:
...
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..... .

Figure 3. We have "true" and "false" regions. What happens in remaining area we don't care as the area doesn't satisfies problem's conditions.
Figure 4. Part of "true/false" regions is gray because it doesn't satisfy x-y<2 condition. Only in "color" part x-y<2 is true.

But in Figure4 we could have doubt about point P: where line x-y<2 passes P, left-above or bottom-right. In first case we will have only "true" region and in second case - "true" and "false" regions. As you correctly pointed out at x=4.5 y=2.5>2.25. So, point P is not included in "color" region (P does not satisfy x-y<2 condition) and x-y<2 passes left-above. So that, we can conclude we have only "true" region.
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Re: Graphic approach to problems with inequalities [#permalink]

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31 Dec 2009, 04:54
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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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02 Jul 2011, 06:28
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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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Re: Graphic approach to problems with inequalities [#permalink]

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28 Jul 2011, 10:14
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Bunuel wrote:
mainhoon wrote:
Can someone solve this without the graphical approach? Bunuel?

My solution from: tough-inequality-challange-89225.html?hilit=walker%20graphic#p732298

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

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

$$\frac{x}{y}>2$$ tells us that $$x$$ and $$y$$ are either both positive or both negative, which means that all points $$(x,y)$$ satisfying given inequality are either in I or III quadrant. When they are both negative (in III quadrant) inequality $$3x+2y<18$$ is always true, so we should check only for I quadrant, or when both $$x$$ and $$y$$ are positive.

In I quadrant, as $$x$$ and $$y$$ are both positive, we can rewrite $$\frac{x}{y}>2$$ as $$x>2y>0$$ (remember $$x>0$$ and $$y>0$$).

So basically question becomes: If $$x>0$$ and $$y>0$$ and $$x>2y>0$$, is $$3x+2y<18$$?

(1) $$x-y<2$$.

Subtract inequalities $$x>2y$$ and $$x-y<2$$ (we can do this as signs are in opposite direction) --> $$x-(x-y)>2y-2$$ --> $$y<2$$.

Now add inequalities $$x-y<2$$ and $$y<2$$ (we can do this as signs are in the same direction) --> $$x-y+y<2+2$$ --> $$x<4$$.

We got $$y<2$$ and $$x<4$$. If we take maximum values $$x=4$$ and $$y=2$$ and substitute in $$3x+2y<18$$, we'll get $$12+4=16<18$$.

Sufficient.

(2) $$y-x<2$$ and $$x>2y$$:
$$x=3$$ and $$y=1$$ --> $$3x+2y=11<18$$ true.
$$x=11$$ and $$y=5$$ --> $$3x+2y=43<18$$ false.

Not sufficient.

Sorry for being naive. Can you please to me the second part of the solution when you conclude it as "Not sufficient". How did you get the numbers 3, 1, 11,5 ?
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Dipankar6435 wrote:
Those were helpful indeed . 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?? ) 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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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Math Expert Joined: 02 Sep 2009 Posts: 38846 Followers: 7720 Kudos [?]: 105948 [1] , given: 11602 Re: Graphic approach to problems with inequalities [#permalink] Show Tags 30 Aug 2013, 06:07 1 This post received KUDOS Expert's post ygdrasil24 wrote: Cool. So does it mean y=f(x) > or < or '=' will decide the region for it . For most of the cases - yes (for example for lines or parabolas). _________________ Manager Joined: 07 Apr 2012 Posts: 126 Location: United States Concentration: Entrepreneurship, Operations Schools: ISB '15 GMAT 1: 590 Q48 V23 GPA: 3.9 WE: Operations (Manufacturing) Followers: 0 Kudos [?]: 11 [1] , given: 45 Re: Graphic approach to problems with inequalities [#permalink] Show Tags 30 Aug 2013, 06:26 1 This post received KUDOS For most of the cases - yes (for example for lines or parabolas).[/quote] One more thing in questions involving greater or less than, a solutions can be infinite. In the attached image when y > x, and x < 1, we can have infinite solutions, right ? if y < x , x < 1 again infinte solutions, right ? So can we safely say if there are 'two' inequalities having less than or greater than condition, we can never have finite solutions , or number of solutions is indeterminate ? Attachments Doc1.pdf [4.25 KiB] Downloaded 38 times  To download please login or register as a user Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7370 Location: Pune, India Followers: 2283 Kudos [?]: 15090 [1] , given: 224 Re: Graphic approach to problems with inequalities [#permalink] Show Tags 11 Aug 2014, 22:17 1 This post received KUDOS Expert's post alphonsa wrote: Can you please tell me how do you know which region to shade? I mean , to the left? or right? above or below the x axis? How do you find out So you draw the line showing the equation represented by the inequality. How do you decide which side of the line does the inequality represent. Usually, you can do that by plugging in (0, 0) in the equation. The point (0, 0) will lie on one side of the line. Put x = 0 and y = 0 in your inequality. If it holds, it means the inequality holds for point (0, 0) and hence will hold for that entire side of the line. So you shade the side where (0, 0) lies. If the inequality does not hold when you put (0, 0), it means (0, 0) is not a solution of the inequality and hence the inequality holds for the opposite side so you shade the opposite side. If the line passes through (0, 0) try any other point which obviously lies on one side of the line. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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

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14 Apr 2015, 22:14
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VeritasPrepKarishma wrote:
erikvm wrote:
I don't understand how we can deduce from (2) that its not sufficient. What I see from the graph, is that y-x=2 never pass the red area. How am I supposed to interpret that graph?

All the other graphs makes sense "Oh okay, don't have any line passing the red part = sufficient" - but the last figure doesn't go through red zone either, yet still insufficient?

Am I supposed to realize that the lines can't pass through that white area just above "TRUE" either?

When the thread gets to the 6th page with, presumably, multiple solutions, you need to quote the post to which you are referring or give its link.

Figure 5. Original post. The line y-x=2 never goes through the red zone - yet "insufficient"

It has nothing to do with whether it passes from the red region or not.

You are given that y - x < 2.
When you draw y - x = 2 in figure 5, you get a line. But what is y - x < 2? It is the area to the right of the line y - x = 2. You can find this by putting in (0, 0) in the inequality y - x < 2. You get 0 - 0 < 2 that is 0 < 2 which is true. Since (0, 0) lies to the right of the line and satisfies the inequality, it means the right of the line is y - x < 2 and the left of the line is y - x > 2.

So you are given that points to the right of y - x = 2 are feasible. Look at all the points to the right of y - x= 2 and that satisfy x/y > 2 (i.e. the green and red regions). Note that of all the points to the right of y - x = 2, some lie in the green region and some lie in the red region (ignoring those that lie in neither region). The ones that lie in green region are those for which 3x+2y < 18. For those that lie in the red region, 3x + 2y > 18. So can we say whether 3x + 2y is less than 18? No. Because some points that satisfy y - x > 2 lie in the reg region and some lie in the green region.

When will an inequality be sufficient. It will be sufficient when the region it points to has points lying only in one region - either red or green.

Mind you, using graphing for inequalities isn't very easy at first. Only once you know instinctively what you are doing will you enjoy it.
Here are three posts that build up on these concepts.
http://www.veritasprep.com/blog/2010/12 ... he-graphs/
http://www.veritasprep.com/blog/2010/12 ... s-part-ii/
http://www.veritasprep.com/blog/2011/01 ... -part-iii/
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Re: Graphic approach to problems with inequalities [#permalink]

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29 Jul 2008, 04:55
Thanks Walker, I use this approach for such problems, but sometime I'm too lazy to make a diagram on computer and post it.. I prefer pen and paper....
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Re: Graphic approach to problems with inequalities [#permalink]

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29 Jul 2008, 06:21
Thnanks for sharing this kind of approach. I will use this approach. Great job.

Thanks again.
Re: Graphic approach to problems with inequalities   [#permalink] 29 Jul 2008, 06:21

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