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Graphic approach to problems with inequalities

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Graphic approach to problems with inequalities [#permalink] New post 29 Jul 2008, 03:51
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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.

see also: http://gmatclub.com/forum/7-t75657

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

Attachment:
tarek99.png
tarek99.png [ 17.04 KiB | Viewed 30836 times ]


see also post by Nach0: Quick Way to Graph Inequalities
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Re: Graphic approach to problems with inequalities [#permalink] New post 29 Jul 2008, 03:55
Thanks Walker, I use this approach for such problems, but sometime I'm too lazy to make a diagram on computer and post it.. :wink: I prefer pen and paper....
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Re: Graphic approach to problems with inequalities [#permalink] New post 29 Jul 2008, 04:02
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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] New post 29 Jul 2008, 05:21
Thnanks for sharing this kind of approach. I will use this approach. Great job.

Thanks again.
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Re: Graphic approach to problems with inequalities [#permalink] New post 30 Jul 2008, 07:44
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 x=4.5 into x-y=2 and find that y=4.25<4.5 (left side).

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.

Additional tip: 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



never thought of using this approach. brilliant!
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Re: Graphic approach to problems with inequalities [#permalink] New post 30 Jul 2008, 07:55
i have no talent in using this, but thanks, good advice!
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Re: Graphic approach to problems with inequalities [#permalink] New post 03 Aug 2008, 08:34
Hi - can someone help me explain Tip #2? I don't really understand what it is saying. Thanks.
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Re: Graphic approach to problems with inequalities [#permalink] New post 03 Aug 2008, 09:27
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bigfernhead wrote:
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] New post 03 Aug 2008, 09:42
Gotcha. Thank you.
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Re: Graphic approach to problems with inequalities [#permalink] New post 03 Aug 2008, 23:49
Very useful, Walker!
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Re: Graphic approach to problems with inequalities [#permalink] New post 04 Aug 2008, 00:42
Is there any other method to solve these type of questions????
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Re: Graphic approach to problems with inequalities [#permalink] New post 07 Mar 2009, 04:40
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.



How do we find which area should be shaded? In first step you defined that set of x and y lies between line x/y=2 and x-axis. Thanks
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Re: Graphic approach to problems with inequalities [#permalink] New post 07 Mar 2009, 08:24
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kbulse wrote:
How do we find which area should be shaded? In first step you defined that set of x and y lies between line x/y=2 and x-axis. Thanks


One of approaches is to check any point.
Let's consider x>0 and try point between x/y=2 and x=0, for example, x=2, y=0.5. For this point x/y=4>2 and the expression is true.
We could also try point above x/y=2 line. For example, x=2, y=2. For this point x/y=1 <2 and the expression is false.
In conclusion, to find what area works, just pick any point in that area (the better to choose a point, for which it is possible to calculate fast).
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Re: Graphic approach to problems with inequalities [#permalink] New post 07 Mar 2009, 09:07
walker wrote:
kbulse wrote:
How do we find which area should be shaded? In first step you defined that set of x and y lies between line x/y=2 and x-axis. Thanks


One of approaches is to check any point.
Let's consider x>0 and try point between x/y=2 and x=0, for example, x=2, y=0.5. For this point x/y=4>2 and the expression is true.
We could also try point above x/y=2 line. For example, x=2, y=2. For this point x/y=1 <2 and the expression is false.
In conclusion, to find what area works, just pick any point in that area (the better to choose a point, for which it is possible to calculate fast).


Hi walker,

how about this way:
x/y>2 => y>x/2 then draw the line y=x/2 and shade everything which is upper side of the line? is it not correct?
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Re: Graphic approach to problems with inequalities [#permalink] New post 07 Mar 2009, 10:05
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kbulse wrote:
how about this way:
x/y>2 => y>x/2 then draw the line y=x/2 and shade everything which is upper side of the line? is it not correct?



1) you should be careful with first operation because it is an inequality and you multiply both sides by y that can have a different sign and change a sign of the inequality.

The correct answer will be:
x/y>2 => x/y * y>2 *y at y>=0 and x/y * y<2 *y at y<0 ==> y<x/2 at y>=0 and y>x/2 at y<0
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Re: Graphic approach to problems with inequalities [#permalink] New post 07 Mar 2009, 10:14
I didn't know that :oops: , i was hesitating whether to ask that question, now I am glad that I asked about that. Thanks
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Re: Graphic approach to problems with inequalities [#permalink] New post 07 Mar 2009, 10:28
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kbulse wrote:
I didn't know that :oops: , 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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Re: Graphic approach to problems with inequalities [#permalink] New post 10 Mar 2009, 00:30
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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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Re: Graphic approach to problems with inequalities [#permalink] New post 10 Mar 2009, 02:34
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Thanks, there is a typo here:
Instead of
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.
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Re: Graphic approach to problems with inequalities [#permalink] New post 10 Mar 2009, 10:06
Yep, it all makes sense to me now..thanks, really appreciate all the effort.
Re: Graphic approach to problems with inequalities   [#permalink] 10 Mar 2009, 10:06
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