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Graphic approach to problems with inequalities

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New post Updated on: 26 Apr 2018, 02:38
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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.

:fyi 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).

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

Image

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: THIS QUESTION

That's all :)
Regards,
Walker

Attachment:
tarek99.png
tarek99.png [ 17.04 KiB | Viewed 55664 times ]

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Originally posted by walker on 29 Jul 2008, 04:51.
Last edited by Bunuel on 26 Apr 2018, 02:38, edited 1 time in total.
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Graphic approach to problems with inequalities  [#permalink]

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New post Updated on: 26 Apr 2018, 03:09
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If \(\frac{x}{y} > 2\), 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.

Not sufficient.


Answer: A.

THIS QUESTION IS DISCUSSED HERE.
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Originally posted by Bunuel on 29 Jul 2008, 05:02.
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Re: Graphic approach to problems with inequalities  [#permalink]

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New post 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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New post 03 Aug 2008, 10:27
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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]

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New post 07 Mar 2009, 05:40
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]

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New post 07 Mar 2009, 09:24
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]

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New post 07 Mar 2009, 10:07
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]

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New post 07 Mar 2009, 11:05
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]

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New post 11 Mar 2009, 11:12
walker wrote:
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.
+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.
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Re: Graphic approach to problems with inequalities  [#permalink]

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New post 11 Mar 2009, 11:34
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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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New post 18 Mar 2009, 20:05
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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Re: Graphic approach to problems with inequalities  [#permalink]

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New post 20 Mar 2009, 13:11
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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.

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


see also post by Nach0: Quick Way to Graph Inequalities



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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New post 13 May 2009, 23:02
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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New post 18 May 2009, 11:24
mdfrahim wrote:
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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New post 02 Aug 2009, 04:53
thanks... but i think it is difficult to use ... ven ur pressed for time
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New post 02 Aug 2009, 10:32
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.
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Re: Graphic approach to problems with inequalities  [#permalink]

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New post 10 Aug 2009, 17:43
Hi Walker - for those of us who are artistically challenged, do you happen to know the way to solve this algebraically? :)
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New post 10 Aug 2009, 23:03
understudy wrote:
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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Re: Graphic approach to problems with inequalities  [#permalink]

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New post 12 Nov 2009, 22:59
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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Re: Graphic approach to problems with inequalities  [#permalink]

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New post 12 Nov 2009, 23:14
babyif19 wrote:
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:


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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Re: Graphic approach to problems with inequalities   [#permalink] 12 Nov 2009, 23:14

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