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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 x-intercept is positive (to the right of the origin)

-x indicates you FLIP EVERYTHING (slope, direction of shading, x-int) -y indicates POSITIVE slope > indicates you shade to the RIGHT (--->) -k indicates the x-intercept is negative

Example: 3x+4y<3

Attachment:

1.gif [ 1.87 KiB | Viewed 20913 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 x-int is to the RIGHT of the origin

-2x+5y<-3

Attachment:

2.gif [ 1.59 KiB | Viewed 20805 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 x-int 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/7-t68037. 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!

+5x: No reversing/reversing/"oppositing" -7y: Positive slope <: shade to the left of the line -4: X-intercept is negative (to the left of the origin)

-3x-6y<-9

Attachment:

4.gif [ 1.84 KiB | Viewed 20580 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 x-intercept. It will be positive because of the -3x

+5x: No reversing/reversing/"oppositing" -7y: Positive slope <: shade to the left of the line -4: X-intercept is negative (to the left of the origin)

-3x-6y<-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 x-intercept. 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.

+5x: No reversing/reversing/"oppositing" -7y: Positive slope <: shade to the left of the line -4: X-intercept is negative (to the left of the origin)

-3x-6y<-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 x-intercept. 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.

wht if k=0? say x-2y>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.

Attachments

Various lines passing thru origin.jpg [ 358.33 KiB | Viewed 20360 times ]

Last edited by mrsmarthi on 09 Mar 2009, 20:09, edited 1 time in total.

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.

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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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.

Attachments

tarek99.png [ 17.7 KiB | Viewed 10626 times ]

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Click below to check out some great tips and tricks to help you deal with problems on Remainders! http://gmatclub.com/forum/compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html#p651942

Word Problems Made Easy! 1) Translating the English to Math : http://gmatclub.com/forum/word-problems-made-easy-87346.html 2) 'Work' Problems Made Easy : http://gmatclub.com/forum/work-word-problems-made-easy-87357.html 3) 'Distance/Speed/Time' Word Problems Made Easy : http://gmatclub.com/forum/distance-speed-time-word-problems-made-easy-87481.html

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