Graphic approach to problems with inequalities

Apologies for asking such a basic question - I'm afraid i haven't yet figured out how to draw these graphs accurately in the actual exam- I've heard you aren't permitted to take a ruler into the center.

How do you draw the figure for eg. x-y < 2 accurately enough to see that the equation passes through point P in your diag. ?

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.


That's all
Regards,
Serg a.k.a. Walker

i have a doubt...while plotting x/y>2...i multiplied both sides with y and got the inequality x>2y .....

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.

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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 ?

Responding to a pm:

To convert x/y > 2 to y = kx form, you can take two cases:

Case 1:
y > 0
x > 2y

Case 2:
y < 0
x < 2y

Now draw x = 2y (see the first diagram of the first post by walker. It's the blue line)
When y > 0 (i.e. region above the x axis), x > 2y. Just plug in a point to figure out the required region. Say, the inequality holds for (4, 1) and hence the required region is below the blue line (but above the x axis)

Similarly, when y < 0 (i.e. region below the x axis), x < 2y. Again, the region between the x axis and x = 2y will be the required region.

So you have got the entire blue region. This is where the relation x/y > 2
For more on plotting regions of inequality, check out this post: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/01 ... -part-iii/

Those were helpful indeed :-D. Thanks. Your Quarter Wit, Quarter Wisdom blog is an eye opener. However i feel that question could have been solved faster by an algebric approach, but only because we were provided that both x and y are positive. In absence of such constraints i think the graphical approach will be faster. Another query though. I found this mentioned at the end of the post. I hope you have come to appreciate the wide range of applicability of graphs. Next time, I will introduce a graphical way of working with Modulus and Inequalities. Could I get a link for that. Thnx in advance
And i faintly remember learning some tricks about plotting graphs in my high school. Something to the tune of-"Graph of y=kx can be drawn by expanding the graph of y=x by k times and y=K+x can be drawn by shifting y=x by k units(to the left or right?? ) Do i need to revisit those for the GMAT??

First of all, hats off, I have tried myself building such approaches but could never get a full proof version that can cover all possible types, urs do.
I have one doubt,

using (i) , x-y<2 will be a region which will pull down the line or the region underneath it ? , So then false condition will also start coming in that region
the region marked grey will come in < condition
using(ii) y-x<2 will be a region which will pull down the line or the region underneath it ? , So then false condition will also start coming in that region which looks ok to me.

Point out if I am interpreting it wrongly.
Thanks.

I have a question, when we say y - x < 2. So we consider region pointing downward as < y condition. But what is the limit of that region.
As per my understanding, its upto infinity, unless crossed out by another line, say x = -5, so our area of consideration gets reduced. Please correct me if I am wrong.
Thanks.

y - x < 2 --> y<x+2. So, it's a whole region below the graph y=x+2 (not limited).

OK. If you refer to the link graphic-approach-to-problems-with-inequalities-68037.html
by Walker, how is the graph x/y >2 drawn here? y < x/2 so graph should be all region less than y = x/2. I didnot get how this graph is drawn.

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.