# Lesson: Inequalities and Absolute Value

## Comment on Inequalities and Absolute Value

### Hi Brent, have a query in the

Hi Brent, have a query in the below question

https://gmatclub.com/forum/if-4-7-x-3-which-of-the-following-must-be-true-168681.html

If 4<(7-x)/3, which of the following must be true?

I. 5<x
II. |x+3|>2
III. -(x+5) is positive

(A) II only
(B) III only
(C) I and II only
(D) II and III only
(E) I, II and III

While I got the equations right, i.e in st2: x<-5 or x>-1 and st 3 is always true, the question asks which of the statement must be true and since st2 yields 2 possible solutions for x, the answer for me is B. Really confused how people have taken st2 to be always true. Great question, Jalaj!

From the given information, we know for certain that x < -5

Given that x < -5, must statement II be true?
From statement II, you determined that EITHER x < -5 OR x > -1
So, if x < -5, must it be true that EITHER x < -5 OR x > -1?
In order to answer YES to the above question we only need one of the parts to be true: x < -5 OR x > -1
Since we can be certain that x < -5.

Notice that the question does NOT read "If x < -5, must it be true that BOTH x < -5 AND x > -1?"

Here's an analogous question:
Let's say you have a rock in your pocket, and someone asks you "Is it true that you have EITHER a rock in your pocket OR an elephant in your pocket?
The answer to that question is YES, it is true that I have EITHER a rock in my pocket OR an elephant in my pocket.

Does that help?

Cheers,
Brent

### Hi Brent, in the question

Hi Brent, in the question below statement III has 2 possible solutions: x < -1 and x > 3.
Therefore, x can have a value of -2, which negates the inequality question |x| > 3.
Therefore my answer is B. Can you help me identify where am I going wrong in this?

If |x| > 3, which of the following must be true?

I. x > 3

II. x^2 > 9

III. |x - 1| > 2

A. I only
B. II only
C. I and II only
D. II and III only
E. I, II, and III

https://gmatclub.com/forum/if-x-3-which-of-the-following-must-be-true-138652-40.html Be careful! We cannot say "statement III has 2 possible solutions x < -1 AND x > 3"
The value of x cannot be less than -1 AND greater than 3 at the same time.
Instead, we can say that EITHER x < -1 OR x > 3

This is very similar to your question above.

I should also point out the error in your comment that "Therefore, x can have a value of -2, which negates the inequality question |x| > 3."
You are approaching this in the wrong direction.
We are told that it is 100% true that |x| > 3, and our job is to identify other statements (from I, II and III) that are 100% true.

Instead, you are assuming that statement III is true and trying to determine whether the given information (|x| > 3) is true.

Does this help?

Cheers,
Brent

### https://gmatclub.com/forum

https://gmatclub.com/forum/the-dark-purple-region-on-the-number-line-above-is-shown-in-its-entire-229498.html
sir how to handle this question? ### Here are two different

Here are two different approaches:

Cheers,
Brent

### https://gmatclub.com/forum/if

https://gmatclub.com/forum/if-x-3-which-of-the-following-must-be-true-138652.html

I am not understanding why statement 3 must be true.

According to the statement x is any number greater than +3 or less than -3. The third choice (which MUST be true) shows that x is any number greater than +3 or less than -1

Doesn't this mean that x could be - 2 which is NOT true? You're reversing the order of the IF-THEN statement.
For example, IF we know that x < 5, THEN it must also be true that x < 8
Likewise, IF we know that x < 0, THEN it must also be true that x < 20

Notice that if we reverse the order, we can something that is false.
For example: IF we know that x < 20, THEN it must also be true that x < 0 (not true!)
This is what you're inadvertently doing.

From the given information, we know that EITHER x > 3 OR x < -3
Let's focus on the fact that x < -3
IF x < -3, THEN can we also be certain that x < -1? YES.

In other words: IF we know that x < -3, THEN it must also be true that x < -1
This is what statement III is saying.

Does that help?

Cheers,
Brent

BTW, here's my solution: https://gmatclub.com/forum/if-x-3-which-of-the-following-must-be-true-13...

### No I am still not following

No I am still not following here.

The way I conceptualised it was looking at the number line:

The first option gives us:

x < - 3

-------- (-3) --------- 0
This is
where x could lie

Compared with:

------ -2 ------ 0
X lies here

OR

So the third choice in the question allows for x = -2.0000000001 which is greater than where x could lie when its x < - 3
Here, x must cannot be anywhere to the right of the number line after -3

That's my reasoning and I still can't seem to fathom how I got it wrong. Have I misinterpreted the number line? ### You're still reversing the IF

You're still reversing the IF-THEN structure of the question.

Here's another analogy of what you are doing.

----------------------------
Consider this question:
Joe is taller than 150 centimeters. Which of the following MUST be true?
i) Joe is taller than 100 centimeters.

Must statement i be true?
Here's how your rationale would look: If Joe is taller than 100 centimeters, then Joe could be 101 centimeters tall, which contradicts the premise that Joe is taller than 150 centimeters.
Therefore statement i is false.

Do you see the problem here?
The given information tells us that Joe is taller than 150 centimeters.
This is 100% true. So, Joe could be 151 centimeters tall, or 152 centimeters tall, or 153 centimeters tall, or 154 centimeters tall, . . . etc.
Also, since Joe is taller than 150 centimeters, we know that Joe CANNOT be 101 cm tall. We also know that Joe CANNOT be 102 cm tall, or 103 cm tall, or 104 cm, or 105 cm etc.

Given that Joe can be 151 cm, 152 cm, 153 cm,...etc, can we be certain that Joe is taller than 100 centimeters?
Yes, absolutely.
So, statement i is true.

----------------------------

Here's one last analogy.
Consider this question:
Joe is a billionaire (i.e., Joe has more than \$1,000,000,000). Which of the following MUST be true?
i) Joe has more than \$1

Must statement i be true?
Here's how your rationale would look: If Joe has more than \$1, then Joe could have \$2, which contradicts the premise that Joe has more than \$1,000,000,000.
Therefore statement i is false.

In actuality, if Joe has more than \$1,000,000,000, then we can be certain that Joe has more than \$1.
So, statement i is true.
------------------------------

Does that help?

Cheers,
Brent ### When you say "So the third

When you say "So the third choice in the question allows for x = -2.0000000001 which is greater than where x could lie when its x < -3", you are reversing the order.

The GIVEN information tells us that x < -3.
So, it could be the case that x = -3.1, or x = -3.2, or x = -4, or x = -5, or x = -5.5, etc
Also, since x < -3, it CANNOT be the case that x = -2.5, or x = -2.0000000001, or x = -2, or x = -1, or x = 0, or x = 4, etc

Given all of this, must it be true that x < -2?
Yes, all of the possible x-values (-3.1, or -3.2, or -4, or -5, or -5.5, etc) are ALL less than -2

Cheers,
Brent

### Ha ha. The second analogy

Ha ha. The second analogy made me see where I went wrong with this one. That and the subsequent explanation. Thanks for this.

### Hi Brent, I would appreciate

Hi Brent, I would appreciate your help on this DS qs:

Is |XY| > X²Y²

1) 0 < X² < 1/4
2) 0 < Y² < 1/9

I understand this property from your lectures-

if |A| > A² ...... then A > A² OR A < -A²

thanks ### Great question!

Great question!
Here's my full solution: https://gmatclub.com/forum/is-xy-x-2-y-2-1-0-x-2-1-4-2-0-y-260816.html#p...

Cheers,
Brent

### Hi Brent,

Hi Brent,

Earlier in the lesson whenever we solved equations containing absolute values we checked if the solutions obtained were actually extraneous roots by substituting the values obtained in the original equation (Concept from Video 25).

Would it not then be necessary to check if the roots obtained extraneous in the above scenario as well? (i.e concepts covered in video 38 and 39)

Thank you in advance for clarifying! Notice that, in my solution, I don't actually solve for x or y. That is, I don't say x must equal some particular value (e.g., x = -1/3). Instead, I made conclusions about certain relationships (e.g., x > x² and y > y²).

As such, I didn't need to check for extraneous roots (since I didn't actually calculate any roots).

Does that help?

Cheers,
Brent

### Yes, Brent! Thank you so much

Yes, Brent! Thank you so much!

### Hi Brent,

Hi Brent,
Can you please explain the rationale behind this statement?
"We know that the absolute value of something will always be greater than or equal to zero"
Thanks ### Consider these examples:

Consider these examples:
|-3| = 3, |7.1| = 7.1, |-4.65| = 4.65, |0| = 0, and |5.06| = 5.06

Notice that, when we find the absolute value of any number, the result is always greater than or equal to zero.

Does that help?

### Hi Brent,

Hi Brent,

Why can we not solve the 1st statement using the same technique as the 2nd? By factoring the eq and getting roots -5 and -3 which means -5<x<-3. This will leave us with only one sol which is -4
Thanks. Good question.

You took the inequality x² - 8x + 15 < 0,
and factored it to get: (x - 3)(x - 5) < 0

How did you conclude that -5 < x < -3?
It should be 3 < x < 5

Cheers,
Brent

### Could you please tell me why

Could you please tell me why statement 3 is correct in the following question:

https://gmatclub.com/forum/if-4-7-x-3-which-of-the-following-must-be-true-168681.html

Thanks,
Aman ### Hi Aman,

Hi Aman,

Here's my full solution: https://gmatclub.com/forum/if-4-7-x-3-which-of-the-following-must-be-tru...

Cheers,
Brent

### Hi Brent,

Hi Brent,

Could you help me please understand how to solve this question?:

https://gmatclub.com/forum/is-x-y-x-y-1-y-x-2-xy-123108.html ### Hi Brent,

Hi Brent,

Great solution,

If I use this approach(find examples to either confirm or deny the statement)

|x - y| > |x| - |y|

x=2
y=1

1>1 FALSE

x=-2
y=1

3>1 TRUE

x=2
y=-1

3>1 TRUE

x=-2
y=-1

1>1 FALSE

So now I can see that the only thing I have to know in order to satisfy the inequality is that x and y have different signs. Because if so, the left side will have an altogether magnitude of two values and the right side will always show the difference in magnitude, no matter what signs are.

In addition, there are cases when you can make the right side negative x=2 y=1, but the left side will always be positive.

Therefore statement one does not tell me anything about the signs of the two values,

Statement two- exactly what I need.

What do you think about my reasoning Brent? ### That's great reasoning.

That's great reasoning.
If x and y have DIFFERENT signs then we can answer the target question with certainty: YES, |x - y| IS greater than |x| - |y|.

I should also mention that, if x and y have the SAME signs, we can also answer the target question of certainty: NO, |x - y| is NOT greater than |x| - |y|.

Cheers,
Brent

### Indeed,

Indeed,

Thank you,

Unfortunately, I always forget about certain denial of the statement which is also considered as a true statement.

e.g.

is x^3>0?

a)x^2>0
b)x<0

If there were such a question the correct answer must have been B, since this is the statement that tells us that it is 100% that x^3>0

Am I right? ### Yes, there are DS questions

Yes, there are DS questions like this, but they are pretty rare.

Cheers,
Brent

### Hi Brent, can you please

https://gmatclub.com/forum/given-the-inequalities-above-which-of-the-following-cannot-be-218973.html ### Brent...last problem in the

Brent...last problem in the video: I am getting x ≥ 0 and x ≤ 4. I test those solutions and they work.

-3x+6 ≥ 6...subtract 6 from both sides...-3x ≥ 0...divide both sides by -3 and I get x ≥ 0.

Where did I go wrong?

Thanks! ### Aside: When solving

Aside: When solving inequalities involving ABSOLUTE VALUE, there are 2 things you need to know:
Rule #1: If |something| < k, then –k < something < k
Rule #2: If |something| > k, then EITHER something > k OR something < -k
(Note: these rules assume that k is positive)

Given: |-3x + 6| ≥ 6
For this question we need to use Rule #2
We get: -3x + 6 ≥ 6 OR -3x + 6 ≤ -6
Let's deal with each inequality separately.

Take: -3x + 6 ≥ 6
Subtract 6 from both sides: -3x ≥ 0
Divide both sides by -3 to get: x ≤ 0
Key concept: When we divide both sides of an inequality by a negative value, we must reverse the direction of the inequality symbol.
In your solution, you forgot to reverse the direction of the inequality symbol.
This is covered in the following lesson: https://www.gmatprepnow.com/module/gmat-algebra-and-equation-solving/vid...

Let's analyze the second inequality...

Take: -3x + 6 ≤ -6
Subtract 6 from both sides: -3x ≤ -12
Divide both sides by -3 to get: x ≥ 4 (I reversed the direction of the inequality symbol)

So our solution is: x ≤ 0 OR x ≥ 4

Does that help?

### Hi Brent,

Hi Brent,
In below question:
https://gmatclub.com/forum/if-4-7-x-3-which-of-the-following-must-be-true-168681.html

I am solving option II as:
-2> x-3 >2 which is giving me x<-5 or x>-1. Since any of these 2 answers can be correct, so I believe that this option II is not sufficient to answer the question i.e. x<-5. Then, why is this option II still correct?
Moreover, ur distance-related explanation looks altogether new and I believe it was no where discussed in ur video-audio chapters, so can u please explain the approach I took to solve?
Thanks. I believe you're reading the question backwards (with regard to what we can assume is true).

We must assume that the given information (4 < (7-x)/3) is true, and then determine which of the three statements must be true.
From the given information, we can conclude that x < -5
In other words, we can be 100% certain that x < -5

For statement II, you're assuming that the inequality |x + 3| > 2 is already true.
If we assume that statement II is true, then you're correct to conclude that EITHER x < -5 OR x > -1.
However, the goal of this question is to determine whether statement 2 is true or not.

Since we can be certain that x < -5, we can also conclude that x + 3 < -2, and if x + 3 < -2, we can be certain that |x + 3| > 2
In other words, we can be certain that statement II must be true.

Does that help?

### https://gmatclub.com/forum/is

https://gmatclub.com/forum/is-x-y-2-x-2-y-243224.html

for this question I chose C because we don't know y and x value for each statement, turns out it is wrong because absolute value is never negative It's a great "fool me once" question. Once you fall for the trap once, you'll likely avoid it the next time you encounter it.

### https://gmatclub.com/forum

https://gmatclub.com/forum/given-the-inequalities-above-which-of-the-following-cannot-be-the-val-218973.html

for this question, oddly I chose A which is wrong for a 50% chance because I already get rid of BCD, could you please explain it? ### Hi Brent, I have a question

Hi Brent, I have a question regarding this problem:

https://gmatclub.com/forum/if-x-is-any-integer-from-6-to-2-inclusive-how-many-227794.html

If we used the same method from the video by using the negative version of the number and then solving, would it be incorrect? With this lower "limit" I get:
0<(x)(x+12)(x-2)<4. But because there's an upper "limit" now of 4, no values from the given values of x from -6 to 2 would make the statement true so that's where I got stuck.

Thank you. Sorry, I should have placed that question under the video that follows the above video. Its solution requires a technique explained in that video (I have moved that link to under the next video)

Having said that, there was a problem with how you derived 0 < (x)(x+12)(x-2) < 4.
So let's discuss that (as well has set the stage for the next video).

In the above video, we learned the following:
1) If |x| < k (where k is positive), then -k < x < k
2) If |x| > k (where k is positive), then z < -k or z > k

Before we can apply one of the above properties, we need to take 2 - |x³ + 10x² - 24x| < 2, and rewrite it in the same form as one of the properties shown above.
In my solution above, I end up with: 0 < |x³ + 10x² - 24x|,
which we can rewrite as follows: |x³ + 10x² - 24x| > 0

At this point, the inequality is in the form |x| > k, which means we can conclude the following:
x³ + 10x² - 24x > 0
or
x³ + 10x² - 24x < 0

It's at this point where you'll need the technique explained in the next video.
When you apply that technique to the two inequalities above, you'll get the same answer.