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inequality graphical approach, what to shade? see inside ...

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inequality graphical approach, what to shade? see inside ... [#permalink] New post 25 Sep 2013, 04:28
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Hey,

I was practicing some of the inequalities questions and saw a lot of you who are using graphical approach to solve inequlity are struggling with which region to shade(or which region is green/true) ..

i just checked this post by nach > quick-way-to-graph-inequalities-76255.html and the way described there is awfully long.

i don't know how to add attachments so will be just writing and using the examples nach used.

1. 3x+4y<3:
take origin as reference .. point ... put it in the above inequality i.e. x=0 and y=0, it'll give you 0<3 which is true hence we'll shade the are which contains origin

2. -2x+5y<-3
take origin as reference and you'll get 0<-3 which is not true .. hence shade the area which does not contain origin .. its that simple ..

refer quick-way-to-graph-inequalities-76255.html for graphical representation of above examples

if a line passes through origin, take 0,1 or 1,0 or -1,0 or 0,-1 as reference point which ever you feel will get you a result faster ..

Hope i helped someone

--
regards,
Stunner
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Manager
Manager
avatar
Joined: 20 Jun 2012
Posts: 96
Location: United States
Concentration: Finance, Operations
GMAT 1: 650 Q50 V28
GMAT 2: 700 Q50 V35
Followers: 1

Kudos [?]: 23 [0], given: 42

GMAT ToolKit User
Re: inequality graphical approach, what to shade? see inside ... [#permalink] New post 25 Sep 2013, 05:56
Here is one example .. I guess I figured out how to add attachments:

Ques. is x negative?

1. x+y=-12
2. y-x<-12

Here it is:

Attachment:
example1.jpg
example1.jpg [ 51.88 KiB | Viewed 562 times ]


1. draw the line x+y=-12
points on this line have both x +ve and -ve .. hence insufficient

2. draw line y-x=-12, put x=y=0(origin as reference) in inequality you'll get 0<-12 which is not true hence shade the region which doesn't contain origin(the dark green one) .. this region also contains value of x both +ve and -ve ..

1 and 2 we get the part of the light green line that is in dark green area and in this region x is +ve, hence sufficient ...
answer is C

Thats a crappy jpg, I know .. :p

will post more of these if anyone would like to see :)
_________________

Forget Kudos ... be an altruist

Manager
Manager
avatar
Joined: 20 Jun 2012
Posts: 96
Location: United States
Concentration: Finance, Operations
GMAT 1: 650 Q50 V28
GMAT 2: 700 Q50 V35
Followers: 1

Kudos [?]: 23 [0], given: 42

GMAT ToolKit User
Re: inequality graphical approach, what to shade? see inside ... [#permalink] New post 25 Sep 2013, 06:34
A question from bunuel's difficult inequality question >> inequality-and-absolute-value-questions-from-my-collection-86939.html

4. Are x and y both positive?
(1) 2x-2y=1
(2) x/y>1

here:
Attachment:
example2.jpg
example2.jpg [ 71.28 KiB | Viewed 555 times ]


1. 2x-2y=1, draw it in x-y plane it passes through three quadrants hence we cannot say both x and y are +ve or not

2. x/y>1, here i took reference point as 0,1 it came 0>1 which is not true, lines which passes through are a bit spcial so we dont just shade the are which doesnt contain the reference point but shade the region between axis and line like I did in above paint-file

1+2 >> both lines are parallel and wont meet far far away in any quadrant .. the light green line in dark green region is the common portion and both x and y will be +ve in this region .. hence answer is C
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Re: inequality graphical approach, what to shade? see inside ...   [#permalink] 25 Sep 2013, 06:34
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Here's a good one, I'd like to see what other approaches kxl19 2 08 Apr 2007, 17:25
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inequality graphical approach, what to shade? see inside ...

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