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

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13 Jan 2012, 00:26

1

This post was BOOKMARKED

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 \(2\)

x=4,y=1 (x/y=4 > 2)

3x+2y = 14 (<18)

x-y = 3 (>2)

x =-5,y=-2

(x/y=2.5>2)

3x+2y= -19 (<18)

x-y = -3 (<2)

So for one case where x-y > 2 and another case where x-y <2 the inequality holds.

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 \(2\)

x=4,y=1 (x/y=4 > 2)

3x+2y = 14 (<18)

x-y = 3 (>2)

x =-5,y=-2

(x/y=2.5>2)

3x+2y= -19 (<18)

x-y = -3 (<2)

So for one case where x-y > 2 and another case where x-y <2 the inequality holds.

Where am I going wrong?

I'm not quite sure what you wanted to illustrated with the numbers you picked but anyway if you want to stick with algebraic solution here it is:

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

\(\frac{x}{y}>2\) 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.

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
_________________

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

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06 Apr 2012, 03:38

Hi Walker ! Great post ! But can you tell me whether there is any way to draw lines on the graph by just seeing the equations? I'm new to this approach and I'm trying to understand it

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

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06 May 2012, 20:56

walker wrote:

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.

Hi walker! I know its been almost 4yrs since you made the original post which has been very helpful to the all here. This approach does appear to ease my handicap with inequalities. Could you please elaborate on your explanation of fig 3 & 4 on how you found the conditions to sufficient or insufficient. As in see even for condition 1 part of the true region in the 1st quadrant is grey?

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

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06 May 2012, 21:06

ok i have tried to answer my question.. pls tell me if i am right: - for 1st condition the area to the left of the line is the region that meets the 1st condition. As can be seen it covers area that is either true or not relevant... but certainly not area that is marked false. - for 2nd equation the area to the right of the line is the region that meets the 2nd condition. As can be seen it covers area that is true, not relevant and ALSO FALSE. And here lies the ambiguity.

So the way to decide is a condition is to ensure that no area covered by it lies in the false region.( Remm it need not all be true but just that it cannot cover any false)

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

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26 Oct 2012, 00:37

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 ?

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.
_________________

“Never Give Up! Never, never, never, never give up. Never!” Winston Churchill

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).
_________________

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

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16 Nov 2012, 07:56

Seems to be a great approach!!!

I was always jumbled up the equations in inequality problems. There is a timing issue since I have just started using it. I am having an issue in finding the exact points of intersection. I mean it's getting solved but requires more time.

But I am hoping that it will improve with practice...

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

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16 Nov 2012, 13:24

Quote:

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.

Been struggling with this point all night ! How do we know that x-y < 2 actually intersects Point P. How ? Is it determined graphically ? or some other means ?

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

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16 Nov 2012, 13:44

sanjay_gmat wrote:

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.

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

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17 Nov 2012, 01:44

tejpreetsingh wrote:

Can you explain me this question with your approach? IS XY>0 1. X-Y>-2 2. X-2Y<-6

@tejpreetsingh

I used the graphical method and got E: both insufficient

Q) encloses 1st & 3rd quadrant completely. 1) Line passing (-2,0),(0,2) slope=+1. Covers only limited part. Hence Insufficient. 2) Line passing (-6,0),(0,3) slope=+1/2. Covers only limited part. Hence Insufficient. Together, Lines intersect at (2,4). Could see the area enclosed between the line not satisfying some area of XY>0 completely. Insufficient.

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.

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

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11 Mar 2013, 00:16

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.

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??

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/

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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