Hi all! My friend Tarek PMed me and asked to show how to use the graphic approach to problems with inequalities. I really love the approach because it isn't only fast one after training, but also reliable. So, I try to illustrate how to use it. :fyi There are two distinct ways to solve math problems: logical and visual. The approach described in this post is for those who use the visual approach. If you like imaging math problems in the visual scratchpad in your mind, you will enjoy this method. At the same time, don't be discourage if it doesn't work for you, try the logical approach instead (described by Bunuel). If \frac{x}{y}>2 , is 3x+2y 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 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. https://gmatclub.com/forum/download/file.php?id=6980 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. See also: THIS QUESTION That's all Regards, Walker tarek99.png

Rewrite x-y x-2. True region is ABOVE (because of ">" sign) the line y=x-2. Rewrite y-x if-x-y-2-is-3x-2y-18-1-x-y-is-less-than-89225.html Hope it helps.

It's not a linear equation. So, the approach is different: MSP12871igh5183c7965eh7000022622h2c27df8334.gif If you are not comfortable with this method use algebra: if-x-y-2-is-3x-2y-18-1-x-y-is-less-than-89225.html

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Graphic approach to problems with inequalities

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Bunuel

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30 Aug 2013, 04:54

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ygdrasil24

Attachment:

tarek99.png

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.

Rewrite x-y<2 as y>x-2. True region is ABOVE (because of ">" sign) the line y=x-2.
Rewrite y-x<2 as y<x+2. True region is BELOW (because of "<" sign) the line y=x+2.

This question is also discussed here: if-x-y-2-is-3x-2y-18-1-x-y-is-less-than-89225.html

Hope it helps.

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13 Sep 2013, 11:45

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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.

Bunuel

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15 Sep 2013, 09:30

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ygdrasil24

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

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15 Sep 2013, 10:27

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Bunuel

ygdrasil24

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.

Bunuel

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15 Sep 2013, 10:48

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ygdrasil24

Bunuel

ygdrasil24

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

It's not a linear equation. So, the approach is different:

walker

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.

Attachment:

MSP12871igh5183c7965eh7000022622h2c27df8334.gif [ 2.71 KiB | Viewed 6673 times ]

If you are not comfortable with this method use algebra: if-x-y-2-is-3x-2y-18-1-x-y-is-less-than-89225.html

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23 May 2014, 00:12

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Bunuel

mainhoon

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...

Bunuel

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23 May 2014, 02:19

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NGGMAT

Bunuel

mainhoon

Can someone solve this without the graphical approach? Bunuel?

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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24 Jun 2014, 04:24

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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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11 Aug 2014, 02:31

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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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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?

KarishmaB

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13 Apr 2015, 21:54

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erikvm

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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VeritasPrepKarishma

erikvm

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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19 May 2016, 00:20

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Bunuel / Chetan2u / Karishma : Your solution to this problem please ?

Thanks
Regards

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26 Apr 2018, 10:36

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Bunuel

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

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26 Apr 2018, 13:12

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gmatacer40

Bunuel

Check the graph below:

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.

Show Spoiler

Attachment:

Graph+.png [ 12.12 KiB | Viewed 3219 times ]

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26 Apr 2018, 20:07

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Bunuel

gmatacer40

Bunuel

Check the graph below:

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.

Show Spoiler

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.

Posted from my mobile device

Bunuel

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26 Apr 2018, 23:28

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gmatacer40

Bunuel

gmatacer40

Check the graph below:

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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27 Apr 2018, 00:55

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gmatacer40

Bunuel

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.

Bunuel

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27 Apr 2018, 03:52

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gmatacer40

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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11 Nov 2018, 07:59

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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.