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Quick Way to Graph Inequalities [#permalink]
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04 Mar 2009, 01:29
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So you're like me and hate inequalities. I've come up with a way to determine what an equality looks like on a graph without plugging in points or solving any math. Most inequalities look like this: ±x±y<±k OR ±x±y>±k where k is a constant This is important to know, because each sign in the equality will tell you something important about the graph. +x indicates you keep EVERYTHING normal (I'll explain later) +y indicates a NEGATIVE slope < indicates you shade to the left (<) +k indicates the xintercept is positive (to the right of the origin) x indicates you FLIP EVERYTHING (slope, direction of shading, xint) y indicates POSITIVE slope > indicates you shade to the RIGHT (>) k indicates the xintercept is negative Example: 3x+4y<3 Attachment:
1.gif [ 1.87 KiB  Viewed 20355 times ]
You look at +3x and know you don't have to do anything with that since it's positive. +4y indicates the slope is NEGATIVE (it's the opposite of +) < indicates you shade LEFT of the line and +3 indicates the xint is to the RIGHT of the origin 2x+5y<3 Attachment:
2.gif [ 1.59 KiB  Viewed 20247 times ]
2x means everything gets REVERSED! Like opposite day... that means +5y (which originally indicates a negative slope) is now a POSITIVE slope and < means you shade to the RIGHT (not the left as usual) and 3 means the xint is POSITIVE (not negative, since x is negative) Got it? It's a little something to memorize, but I believe it would help a lot in combination with Walker's Graphic Approach located here: http://gmatclub.com/forum/7t68037. I think it would definitely save a lot of time when working on some DS problems, which are KNOWN to have tons of inequalities. Tell me what you think. This is my first post, so go easy on me I can do more examples if you need! Nach0 wrote: Here's a link where you can test out what you've just learned: http://wims.unice.fr/wims/en_H6~analysi ... eq.en.htmlIn "Type of Region" select Linear I or Linear II. It will either provide you with a graph or an inequality. From then you must match it with the corrersponding answer. You can practice taking what you've learned and applying right away. It will help to quickly determine what kind of grap you are working with. If you're looking for a tool to create graphs as an image, you can use this website: http://www.hostsrv.com/webmab/app1/MSP/ ... s&s3=basicJust plug in the inequation/equation and generate. Then save your pic!
Last edited by Nach0 on 04 Mar 2009, 06:59, edited 1 time in total.



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Re: Quick Way to Graph Inequalities [#permalink]
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04 Mar 2009, 04:21
Wow! dat was an amazing post. u get +1 in ur first post itself from me pls post sum more examples



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04 Mar 2009, 06:28
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Here's a couple more examples: 5x7y<4 Attachment:
3.gif [ 1.91 KiB  Viewed 20030 times ]
+5x: No reversing/reversing/"oppositing" 7y: Positive slope <: shade to the left of the line 4: Xintercept is negative (to the left of the origin) 3x6y<9 Attachment:
4.gif [ 1.84 KiB  Viewed 20024 times ]
3x: Everything will be reversed! 6y: Positive slope, but since there's a 3x, it will be negative <: Supposed to mean shade to the left, but it's right this time because of the 3x 9: Like the others, generally means negative xintercept. It will be positive because of the 3x



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Re: Quick Way to Graph Inequalities [#permalink]
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04 Mar 2009, 06:40
Nach0 wrote: Here's a couple more examples: 5x7y<4 Attachment: 3.gif +5x: No reversing/reversing/"oppositing" 7y: Positive slope <: shade to the left of the line 4: Xintercept is negative (to the left of the origin) 3x6y<9 Attachment: 4.gif 3x: Everything will be reversed! 6y: Positive slope, but since there's a 3x, it will be negative <: Supposed to mean shade to the left, but it's right this time because of the 3x 9: Like the others, generally means negative xintercept. It will be positive because of the 3x Excellent +1 from me. the exact values can be plotted by substituting x =... 2,1.0,1,2,3.... which will be useful in merging two graphs and finding the intersection.



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Re: Quick Way to Graph Inequalities [#permalink]
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04 Mar 2009, 06:50
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selvae wrote: Nach0 wrote: Here's a couple more examples: 5x7y<4 Attachment: 3.gif +5x: No reversing/reversing/"oppositing" 7y: Positive slope <: shade to the left of the line 4: Xintercept is negative (to the left of the origin) 3x6y<9 Attachment: 4.gif 3x: Everything will be reversed! 6y: Positive slope, but since there's a 3x, it will be negative <: Supposed to mean shade to the left, but it's right this time because of the 3x 9: Like the others, generally means negative xintercept. It will be positive because of the 3x Excellent +1 from me. the exact values can be plotted by substituting x =... 2,1.0,1,2,3.... which will be useful in merging two graphs and finding the intersection. Also kind of fine tuning the logic (still you hold the patent ) for the scenario 2, where you said when X coeff. is negative, can be converted to scenario 1 by multilying 1 both the side. So 4x+5y<6 becomes, 4x 5y >6 the diagram still look the same. and the user dont have to remember too much of logics.



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Re: Quick Way to Graph Inequalities [#permalink]
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04 Mar 2009, 06:53
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Here's a link where you can test out what you've just learned: http://wims.unice.fr/wims/en_H6~analysi ... eq.en.htmlIn "Type of Region" select Linear I or Linear II. It will either provide you with a graph or an inequality. From then you must match it with the corrersponding answer. You can practice taking what you've learned and applying right away. It will help to quickly determine what kind of grap you are working with.



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Re: Quick Way to Graph Inequalities [#permalink]
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07 Mar 2009, 03:22
wht if k=0? say x2y>0 How will the graph be like?



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07 Mar 2009, 05:16
2 Nach0
how do you find at which points lines intersect axises? thanks



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07 Mar 2009, 10:12
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ritula wrote: wht if k=0? say x2y>0 How will the graph be like? Ritula, When k = 0, then the striaght line will always pass thru the origin. The way that I do is, make it if the form y<=>= mx. Depending on whether m(which is the slope) is positive or negative, we determine how we can draw the graph. If m is positive, the line runs from bottom left to top right. If m is negative, the line runs from top left to bottom right. One more thing that you can make a note. If m < 1(absolute value of slope) then line is inclined towards X axis. If m > 1(absolute value of slope) then line is inclined towards Y axis. If m = 1(absolute value of slope) then line is inclined exactly 45 degrees with X axis. In our case, x  2y>0, can be written as y<1/2x. Slope is positive, and is less than 1. Hence the graph looks like the below. I have drawn the cases where the slope is positive and negative and also when the slope is > 1 or less than 1. Hope you get the idea now.
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Various lines passing thru origin.jpg [ 358.33 KiB  Viewed 19803 times ]
Last edited by mrsmarthi on 09 Mar 2009, 20:09, edited 1 time in total.



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07 Mar 2009, 10:22
kbulse wrote: 2 Nach0
how do you find at which points lines intersect axises? thanks This is simple. Put x = 0, and find the y intercept(treating inequality as equality sign). Then the point obtained will be (0, y intercept) which is the point on Y axis. Similary put x =0 and find the x intercept(treating inequality as equality sign). Then the point obtained will be (x intercept,0) which is the point on X axis.



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09 Mar 2009, 15:20
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Yes Mrs Marthi is correct. I actually found her post on inequalities very useful in combination to mine. If 5X2Y>3 then to find the X axis, just take the 5 from 5X (because you want the Xaxis) and have 3 (or k) divide by 5. (3/5,0) To find the Y axis, just take the 2 (because you want the Yaxis) and have 3 (or k) divide by 2. (0,3/2) Basically, take the coefficient from the letter corresponding to the axis you want and have the kconstant divide it. You want Xaxis: take the number in front of x and divide it from K same goes with the Yaxis Attachment:
5.gif [ 1.9 KiB  Viewed 19699 times ]
Hope this helps... correct me if I'm wrong.



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31 May 2009, 08:34
Thanks a lot.. very useful indeed



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01 Jun 2009, 16:48
It is useful indeed !



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03 Jun 2009, 11:19
Thanks a lot.. very useful
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Re: Quick Way to Graph Inequalities [#permalink]
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03 Jun 2009, 11:38
Nach0 wrote: Here's a link where you can test out what you've just learned: http://wims.unice.fr/wims/en_H6~analysi ... eq.en.htmlIn "Type of Region" select Linear I or Linear II. It will either provide you with a graph or an inequality. From then you must match it with the corrersponding answer. You can practice taking what you've learned and applying right away. It will help to quickly determine what kind of grap you are working with. Thanks Nach0 for the link above. Really very helpful. Do you know if there are any other exercises on this serevr which may be good for GMAT?
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Re: Quick Way to Graph Inequalities [#permalink]
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01 Jul 2009, 05:57
mrsmarthi: for the equation x2y>0 how do we know which side to shade?kindly explain.
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22 Jul 2009, 22:02
Thank you very much!!!



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22 Jul 2009, 23:56
Thanks, very useful.



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07 Nov 2009, 08:26
Excellent explanation! Thanks!



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10 Nov 2009, 16:15
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This is a very good post. Quick solving is the key in GMAT and this certainly goes a long way in helping! There is one thing I would like to add though. Suppose we encounter an equation in which we are required to multiply or divide by the variable x or y (as in the question explained by walker in his topic dealing with graphs), then we have to simply break it up into two cases and solve it. Let me take the following equation as an example (those of you who have gone through walkers post might recognize this equation): \((x/y) > 2\) now in order to plot it using the 'quick' approach, we have to break it up in to two cases. 1) When y is positive: \(x > 2y\) which can be written as \(x  2y > 0\) This case gives us the required line. However, it only gives us the region for y>0. To find the region for y<0, we need case 2. 2) When y is negative: \((x/y) > 2\) , now multiply both sides with 1 and simplify. we get \(x + 2y < 0\) This case gives us the region for all y<0. Thus from the above two cases, we infer that: 1) For all positive values of y, the region lies to the right (as given by the equation in case 1). 2) For all negative values of y, the region lies to the left (as given by the equation in case 2). Note: x = 0 is the line above which all y is positive and below which all y is negative. Thus, it will act as a boundary line. Also, while plotting the line, we should think of an '=' present instead of the inequality in the original equation. The inequality comes into the picture when we want to plot the region.
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tarek99.png [ 17.7 KiB  Viewed 10070 times ]
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