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Graphic approach to problems with inequalities

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

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New post 22 May 2014, 23:12
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.

Answer: A.


Hi Bunnel

I understood the part that x & y are either both positive or both negative.... when both are positive we get x>2y.... then we solve on that basis... what happens to when both are negative???? we solved both statements based on this and also declared statement 1 sufficient.
when both x & y are negative then we cannot have x>2y right? so where are we considering this...
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Re: Graphic approach to problems with inequalities  [#permalink]

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New post 23 May 2014, 01:19
NGGMAT wrote:
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.

Answer: A.


Hi Bunnel

I understood the part that x & y are either both positive or both negative.... when both are positive we get x>2y.... then we solve on that basis... what happens to when both are negative???? we solved both statements based on this and also declared statement 1 sufficient.
when both x & y are negative then we cannot have x>2y right? so where are we considering this...


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

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New post 24 Jun 2014, 03:24
Can we use this technique to solve any of the in equality ques. My accuracy in such questions suffer.
Please mention how I should approach in equalities.....
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New post 24 Jun 2014, 05:00
GmatDestroyer2013 wrote:
Can we use this technique to solve any of the in equality ques. My accuracy in such questions suffer.
Please mention how I should approach in equalities.....


Graphic approach is not a silver bullet for inequality questions. Yes, it's handy for some of them but certainly not for all of them.

Hope the links below help to attack them more efficiently:

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

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New post 11 Aug 2014, 01:31
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 :?:
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Re: Graphic approach to problems with inequalities  [#permalink]

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New post 11 Aug 2014, 21:17
1
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.
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Re: Graphic approach to problems with inequalities  [#permalink]

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New post 13 Apr 2015, 00:16
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?
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New post 13 Apr 2015, 20:54
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.
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New post 14 Apr 2015, 00:29
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"
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New post 14 Apr 2015, 21:14
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erikvm wrote:
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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New post 18 May 2016, 23:20
Bunuel / Chetan2u / Karishma : Your solution to this problem please ?


Thanks
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New post 19 May 2016, 21:06
parvgugnani wrote:
Bunuel / Chetan2u / Karishma : Your solution to this problem please ?


Thanks
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You can also consider number line.

1. x - y < 2

________________________y __ __ x____________

x cannot be 2 units or more to the right of y. It is either closer or to the left of y (in the green region)
But x/y > 2 so x is more than twice of y.
If y is positive, y must be less than 2. So x will be less than 4 (cannot be 2 or more away from y).
y can take any negative value. x would be to the left of y. When x is to the left of y, it can take any value.
(If say y is 1, x could be 2.5; If y is -1, x could be -2.5)

Is 3x + 2y < 18?
x is less than 4 and y is less than 2 so 3x + 2y is less than 3*4 + 2*2 i.e. 16.
Sufficient.

2. y - x < 2

_________________________x __ __ y ___________

y cannot be 2 units or more to the right of x. It is either closer or to the left of x. (in the green region)
But x/y > 2 so x is more than twice of y.
If y is positive, any value of y is acceptable. y will be to the left of x.

Is 3x + 2y < 18?
We immediately note that since x and y can take large positive values or small ones, 3x + 2y may or may not be less than 18. Not sufficient.

Answer (A)
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Re: Graphic approach to problems with inequalities  [#permalink]

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New post 26 Apr 2018, 09:36
Bunuel wrote:
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.


Bunuel I have one doubt. if we see the algebraic approach, the the first condition gives us the solution as y<2 and x<4. However, if i see the graphical approach, these values doesn't satisfy the graphical area satisfied by original condition and condition-1. Clarify where am I going wrong?
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New post 26 Apr 2018, 12:12
gmatacer40 wrote:
Bunuel wrote:
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.


Bunuel I have one doubt. if we see the algebraic approach, the the first condition gives us the solution as y<2 and x<4. However, if i see the graphical approach, these values doesn't satisfy the graphical area satisfied by original condition and condition-1. Clarify where am I going wrong?


Check the graph below:
Image

Blue region satisfies the following conditions:
\(x>0\) and \(y>0\) and \(x>2y>0\)
\(x-y<2\).

As you can see for that region \(y<2\) and \(x<4\) holds true.


Attachment:
Graph+.png
Graph+.png [ 12.12 KiB | Viewed 858 times ]

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

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New post 26 Apr 2018, 19:07
Bunuel wrote:
gmatacer40 wrote:
Bunuel wrote:
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.


Bunuel I have one doubt. if we see the algebraic approach, the the first condition gives us the solution as y<2 and x<4. However, if i see the graphical approach, these values doesn't satisfy the graphical area satisfied by original condition and condition-1. Clarify where am I going wrong?


Check the graph below:
Image

Blue region satisfies the following conditions:
\(x>0\) and \(y>0\) and \(x>2y>0\)
\(x-y<2\).

As you can see for that region \(y<2\) and \(x<4\) holds true.


Attachment:
Graph+.png


Bunuel
Bunuel, what if we select values x=3.9 and y=0.1.
It satisfies x>0 y>0 x<4 and y<2. However, I observe that this point doesn't come in graphical solution.please explain.

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

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New post 26 Apr 2018, 22:28
gmatacer40 wrote:
Bunuel wrote:
gmatacer40 wrote:
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.


Bunuel I have one doubt. if we see the algebraic approach, the the first condition gives us the solution as y<2 and x<4. However, if i see the graphical approach, these values doesn't satisfy the graphical area satisfied by original condition and condition-1. Clarify where am I going wrong?


Check the graph below:
Image

Blue region satisfies the following conditions:
\(x>0\) and \(y>0\) and \(x>2y>0\)
\(x-y<2\).

As you can see for that region \(y<2\) and \(x<4\) holds true.

Bunuel
Bunuel, what if we select values x=3.9 and y=0.1.
It satisfies x>0 y>0 x<4 and y<2. However, I observe that this point doesn't come in graphical solution.please explain.

Posted from my mobile device


I see what you mean.

x < 4 and y < 2 does NOT mean that ALL points which have have x-coordinate less than 4 and y-coordinate less than 2 will satisfy \(x>0\), \(y>0\), \(x>2y>0\) and\(x-y<2\).

It meas that ALL points which satisfy \(x>0\), \(y>0\), \(x>2y>0\) and\(x-y<2\) will have x-coordinate less than 4 and y-coordinate less than 2.

Or in another way, ALL points in blue region have x-coordinate less than 4 and y-coordinate less than 2 but NOT all points which have x-coordinate less than 4 and y-coordinate less than 2 are in blue region.

Do you see the difference? Does this make sense?
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Re: Graphic approach to problems with inequalities  [#permalink]

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New post 26 Apr 2018, 23:55
gmatacer40 wrote:
Bunuel wrote:
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.


Bunuel I have one doubt. if we see the algebraic approach, the the first condition gives us the solution as y<2 and x<4. However, if i see the graphical approach, these values doesn't satisfy the graphical area satisfied by original condition and condition-1. Clarify where am I going wrong?


Bunuel
I see. So, as per our discussion, I conclude that algebraically method of solving has a limitation in the way that it provides superset of solution unlike graphical approach, which gives precisely the solution set only. Am I correct?
However, it seems weird to me since I used to see inequality algebraic method in the other way. For example, if x>5 and y<2 that implies x-y>3; the solution is so precise that every point greater than 3 satisfies the equation. So, still not very clear.
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Re: Graphic approach to problems with inequalities  [#permalink]

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New post 27 Apr 2018, 02:52
gmatacer40 wrote:
Bunuel
I see. So, as per our discussion, I conclude that algebraically method of solving has a limitation in the way that it provides superset of solution unlike graphical approach, which gives precisely the solution set only. Am I correct?
However, it seems weird to me since I used to see inequality algebraic method in the other way. For example, if x>5 and y<2 that implies x-y>3; the solution is so precise that every point greater than 3 satisfies the equation. So, still not very clear.


Not entirely. The algebraic way, gives you precise ranges too. The point is that for (1) we not only have that x < 4 and y < 2, we also have the other constraints: \(x>0\), \(y>0\), \(x>2y>0\) and\(x-y<2\). If you apply all of them you'll get the same domain as we got from the graphic approach, namely the blue region in my post above.
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New post 11 Nov 2018, 06:59
Hi, I have an alternate approach for the same. Please correct if Im wrong.

Given: x/y > 2
=> Inference:
a) x,y are both +ve or both are -ve
b) |x| > |y|

Now,
(1) x - y < 2
if both are positive, then , let x = 7, y = 1
7 - 1 is not LT 2
if both are negative, then, let x = -7, y = -1
-7 - (-1) => -6 < 2 (True)
hence both x and y are negative
therefore in the eq 3x + 2y => always < 0 => sufficient

(2) y - x < 2
if both positive, let x = 7, y = 1, 1 - 7 = -6 < 2 (True)
if both negative, let x = -7, y = -1, -1 - (-7) => 7 not LT 2
Hence both are positive
therefore, if positive values are put in the inequality 3x + 2y => clearly insufficient

Hence, answer is A.
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