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# Why is this simple inequality not giving stable answers?

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Joined: 06 Sep 2018
Posts: 37
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27 Sep 2018, 02:10
Hi all! Thanks for your help in advance. This basic question is driving me crazy! Imagine that I have two inequalities,let's call them A and B.

A is (2 < x)
B is (y + 3< x).

My understanding is that if I merge these two inequalities, I am going to get a new inequality C that will yield valid solutions for itself and also the previous two inequalities. My understanding is that to merge A and B, I can only add them up when the signs face the same direction. I did that and got a new inequality C which is

C is (y + 5 < 2x).

Now, if x is 2.5 (which is greater than 2 hence meets inequality A), then substituting 2.5 for x on C gives me y<0 HOWEVER substituting in B gives me y<-0.5. Why? Shouldn't both give me the same?

Thanks a lot.
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27 Sep 2018, 08:56
1
Hi gmat800live,

this is an interesting problem. I might be wrong, but common sense tells me that y should be less than -1 (solved it graphically). Maybe Bunuel could shed some light on this problem?
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27 Sep 2018, 09:24
1
Hi gmat800live,

In simple terms, the reason why the 'solution' to the two inequalities changes is because you are NOT adding the same value to each side of larger inequality.

Here's a simple example:

0 < X

IF we add 1 to BOTH sides, we have...

(0+1) < (X+1)
1 < X + 1

The solution set for the two inequalities is the SAME because those 2 inequalities are the SAME (once you simplify them).

In your example, you are NOT adding the same value to both sides. You're adding '2' to the 'left side' and 'X' to the 'right side', so you are fundamentally changing the inequality (it's DIFFERENT from the one you started with) - thus, the solution sets will be different.

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27 Sep 2018, 10:06
1
gmat800live wrote:
Hi all! Thanks for your help in advance. This basic question is driving me crazy! Imagine that I have two inequalities,let's call them A and B.

A is (2 < x)
B is (y + 3< x).

My understanding is that if I merge these two inequalities, I am going to get a new inequality C that will yield valid solutions for itself and also the previous two inequalities. My understanding is that to merge A and B, I can only add them up when the signs face the same direction. I did that and got a new inequality C which is

C is (y + 5 < 2x).

Now, if x is 2.5 (which is greater than 2 hence meets inequality A), then substituting 2.5 for x on C gives me y<0 HOWEVER substituting in B gives me y<-0.5. Why? Shouldn't both give me the same?

Thanks a lot.

If you add together two inequalities that are both true, and the signs are pointing the same way, you'll get a third inequality that's also true.

For example, if x < 10, and y < 3, you can be completely certain that x + y < 13.

The problem you're running into is that it doesn't go the other way. If x + y < 13, that doesn't mean that x < 10 and y < 3. (You can see this by testing numbers, just like you did - x could be 11 and y could be 1, in which case the first of those inequalities is false!)

This is what you're doing in your example, because you're starting with the 'new' inequality you got after doing the addition, and then trying to apply the results you get from it to the 'old' inequalities you added together. That doesn't necessarily work.

In short:

- Adding two correct inequalities together will give you a new, correct inequality
- However, 'splitting up' one correct inequality won't always give you two correct inequalities
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28 Sep 2018, 00:34
Thank you so much Akela, EMPOWERgmatRichC and ccooley for looking into this! This has opened some sort of pandora's box :D! I realize that my understanding of some basic concepts is not strong as I thought but I am extremely happy I finally got it! Makes total sense what you guys have said and it has made me think a lot! Final summary is that when I say:

A: 2 < x
B: y + 3 < x

And I add them up to get C: y + 5 < 2x. Then the statement that is true is:

If you give me values of x and y that satisfy A AND B, I can assure you, that they will also meet C

That's it . There will then be values that meet for instance C and B, but not A etc! But those shouldn't distract us from the true valid statement above. Gosh... it took me time to get here, but glad it happened. Thanks guys and thank @willteiss for being part of this small discovery :D!
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28 Sep 2018, 01:53
If you give me values of x and y that satisfy A AND B, I can assure you, that they will also meet C

Yes, this is the right conclusion you drawn. And you will see it used in making of so many DS questions.

When timer is running we tend to forget CHECKING the value of a variable, by putting it into the original inequalities, whether it is VALID.
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Re: Why is this simple inequality not giving stable answers? &nbs [#permalink] 28 Sep 2018, 01:53
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# Why is this simple inequality not giving stable answers?

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