# Lesson: Inequalities - Part I

## Comment on Inequalities - Part I

Good point!! The question is somewhat ambiguous.
The intent of the question is to compare the lengths of line segments (not to compare the products AB and BC)

To avoid this ambiguity, an official GMAT question would add some words to explain this. For example:
"Is LINE SEGMENT AB longer than LINE SEGMENT BC?"

Cheers,
Brent

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

https://gmatclub.com/forum/is-x-3-2-1-2-3-x-1-x-3-2-x-x-62992.html

For this question I get that : Rephrase to |x-3| = 3-x?
1) x-3 = 3-x and x = 3
2) -(x-3) = 3-x , so -x+3 = 3-x and x=anything

How do you then get to the step that x <= 3? apart from thinking about the range from (2)

Here's how I'd look at this question:
Rewrite right side as follows: Does |x-3| = -(x-3) ?

At this point, we should recognize that |x-3| is always greater than or equal to zero.
In other words, |x-3| ≥ 0
So, we need it to be the case that -(x-3) ≥ 0
Multiply both sides by -1 to get: (x-3) ≤ 0
Add 3 to both sides to get: x ≤ 3

So, in order for |x-3| to equal -(x-3), it must be the case that x ≤ 3

Does that help?

Cheers,
Brent

### Hi Brent,

Hi Brent,

https://gmatclub.com/forum/is-x-2-1-x-239236.html

I understood you explanation. However, I still have a doubt

I concluded that, The target question will be true only if X will be an integer. However, in statement 2, the inequality is true for all integer values but it is false for any decimal value. I am not sure where I am going wrong. So according to me Statement B is insufficient.

Pls help

Thanks,
Sushant

Target question: Is x² > 1/x ?

"I concluded that, The target question will be true only if X will be an integer"
This isn't correct.
For example, if x = -1.5 (non-integer), we can see that x² > 1/x
Conversely, if x = 1 (integer), we get x² = 1/x (not x² > 1/x)

Does that help?

Cheers,
Brent

### Hi Brent,

Hi Brent,

https://gmatclub.com/forum/if-x-and-y-are-nonzero-integers-is-x-1-y-108741.html

### Until 02:48 will inequality

Until 02:48 will inequality remain even if we do root or a power to both sides of inequality?

### That's correct. However, it's

That's correct. However, it's correct ONLY when both quantities are positive.
That is, if x and y are both positive and x < y, then it must also be the case that:
1) √x < √y
and 2) x² < y²

Conversely, if x and y are NOT both positive, we can't generalize anything about the roots and powers.

For example, it's true that -5 < 3
However, when we SQUARE both sides we get: 25 < 9, which is not true.

Likewise, if we take the inequality -9 < 4
And try to find the SQUARE ROOT of both sides, we got stuck because √(-25) does not have a real value.

Does that help?

### It helped greatly in

It helped greatly in understanding the concept with a detailed explanation.

### Thanks for such detailed

Thanks for such detailed teaching, I was wondering, what happens when we Square both the sides of an inequality or when the square root or just root both sides of an inequality, (as we do in an equation)?

### Good question!

Good question!

SQUARING BOTH SIDES:
- If 0 < x < y, then 0 < x² < y²
- If x < 0 < y, then we need extra information to compare x² and y²
- If x < y < 0, then y² < x²

SQUARE-ROOTING BOTH SIDES:
- If 0 < x < y, then 0 < √x < √y
- If x < 0 < y, then √x has no defined value
- If x < y < 0, then √x and √y have no defined value

Cheers,
Brent

### Hi Brent,

Hi Brent,
Pls can you explain me this problem
Is a > |b|?
(1) 2^(a-b) > 16
(2) |a - b| < b

### Tricky!!

Tricky!!
Here's my full solution: https://gmatclub.com/forum/is-a-b-198568-20.html#p2498608

Cheers,
Brent

### Hi Brent,

Hi Brent,
Is x > 0?
(1) (x^2 - 1)(x^3) > 0
(2) x^2 < 1

### Hi Brent, while working with

Hi Brent, while working with the below question - statement one, it is not possible to explore taking a squre root on both sides? Thanks.

Is x > 0?

(1) (x + y)^2 < (x - y)^2
(2) x + y < x - y

### Great question.

Great question.
For inequalities, taking the square root of both sides only works if both values (before they're squared) are positive.
So, for example, if x and y are both positive and x² < y², then we can conclude that x < y.

However, the property doesn't apply if x or y are negative.
For example, we know that (1)² < (-2)², but we can't then conclude that 1 < -2

Cheers,
Brent

### Many Thanks Brent for the

Many Thanks Brent for the explanation.

### Hi Brent,

Hi Brent,

Not sure if this is the ideal section for this inequality section, but i'm hoping you can guide me in answering the following question.

https://gmatclub.com/forum/if-xy-0-is-x-3-y-268848.html

If you multiply the cube root of both sides of the inequality at the beginning it gives you X + Y > 0. Which perfectly coincides with A being sufficient to answer. Is my approach valid in simplifying the target inequality at the beginning?

Thanks,
Marc

That's a great idea, Marc. However, the cube root of x³ + y³ isn't x + y

Here's why:
The cube root of 8 is 2, since 2³ = 8
Likewise, the cube root of 125 is 5, since 5³ = 125

So, for the cube root of x³ + y³ to be x + y, it must be the case that (x + y)³ = x³ + y³

Let's expand and simplify (x + y)³
We get: (x + y)³ = (x + y)(x + y)(x + y)
= (x + y)(x² + 2xy + y²)
= x³ + 3xy² + 3x²y + y³

Since (x + y)³ = x³ + 3xy² + 3x²y + y³, we know that (x + y) is actually the cube root of x³ + 3xy² + 3x²y + y³

The great thing is that, even though the cube root of x³ + y³ isn't x + y, this mistake works perfectly, house since statement 1 is, indeed, sufficient :-)

### Hi Brent, in this question -

Hi Brent, in this question - https://gmatclub.com/forum/if-mn-0-and-k-m-l-n-mn-which-of-the-following-must-be-true-273169.html

You have rewritten the fraction as Combined fractions: (kn+lm)/mn<mn

MY QUESTION is, can you take the LCM of a fraction like that without knowing if the fractions are positive or negative in nature?

Great question.

The key here is that we don't necessarily need the LCM to have common denominators.
For example, if we want to add 1/6 + 1/4, we don't have to rewrite the fractions with denominator 12 (which is the LCM of 6 and 4)
We can also just go with the denominator that's equal to the product of 6 and 4.
So, 1/6 + 1/4 = 4/24 + 6/24 = 10/24, which simplifies to be 5/12

Likewise, to add k/m + l/n, we can use the common denominator mn.
Notice, that if m is positive and n is negative, the common denominator mn will still be the same for both fractions, and that's all that matters when it comes to adding (or subtracting) fractions.

### Hi Brent,

Hi Brent,

https://gmatclub.com/forum/if-abcd-is-not-equal-to-zero-is-abcd-215779.html

### Thanks Brent, I did

Thanks Brent, I did understand your second approach better.

However, I'm not sure how I can finish this question in 2 mins :( Would you say this is a 700 level question?

### Yes, this is most definitely

Yes, this is most definitely a 700+ level question.

### Need help with this question

Need help with this question stem:
https://gmatclub.com/forum/if-x-and-y-are-positive-integers-is-x-y-x-5-y-193981.html

I got answer D. For statement 1, I substituted y=5 and got x<5. Therefore, x is not equal to y and statement 1 is sufficient.

What is incorrect with the above reasoning? Thanks!

Be careful. You're treating the target QUESTION as though it's a true statement.

Target question: Is x/y < (x+5)/(y+5)?
(1) y = 5
If y = 5, then the target question becomes: Is x/5 < (x+5)/(5+5)?
Simplify: Is x/5 < (x+5)/10?
Multiply both sides of the inequality by 10 to get: Is 2x < x + 5?
Subtract 5 from both sides to get: Is x < 5?
Since we don't know the value of x, there's no way to answer the resulting target question: "Is x < 5?"
As such, statement 1 is not sufficient.

### Ok, I got it now. We were

Ok, I got it now. We were missing the relationship of x with regards to y in the original question, so ST1 in insufficient as we only have value for y.
Thank you!

Exactly!

### Hi Brent,

Hi Brent,
In this question https://gmatclub.com/forum/is-xy-6-1-x-3-and-y-2-2-1-2-x-2-3-and-y-85712.html

the target question asked if xy < 6.
in the (1) statement, we say x < 3 and y < 2
so I multiplied xy < 6 and felt it was sufficient.

I did get the answer wrong, but would it be fair to say that only what is stated by each statement is true, but any inferences are not necessarily true?
so x < 2 and y < 3 is the only thing we can consider and not xy < 6.

almost sounds like critical reasoning.

I am revisiting inequalities as it is a weak area.

Good question. In the second video about inequalities (https://www.gmatprepnow.com/module/gmat-algebra-and-equation-solving/vid...), you'll see that it's okay to ADD inequalities if the inequality symbols are facing the same direction.
So for this question, we COULD conclude that x + y < 5

The problem with multiplying both values is that either value (x or y) can be positive or negative, and if both values are negative, then the product will be positive.
For example, it could be the case that x = -10 and y = -10, in which case xy = (-10)(-10) = 100, which is definitely NOT less than 6.

### Hi Brent,

Hi Brent,

How would you solve this question?

https://gmatclub.com/forum/if-n-denotes-a-number-to-the-left-of-0-on-the-number-line-such-that-91659.html

### Hi Brent, I have a question

Hi Brent, I have a question regarding this question:

https://gmatclub.com/forum/if-mn-0-and-k-m-l-n-mn-which-of-the-following-must-be-true-273169.html

After combining the fractions to (kn+lm)/mn < mn, since it's given that mn <0, can we then say that (kn+lm)/mn < 0? And because mn is negative, then we would have to reverse the equality sign when multiplying so it becomes (kn+lm) > 0. Finally, I thought since mn <0, squaring it would make it >0 since any (negative)^2 is positive, so I chose choice c. However, for this final step, I'm not sure if it's a reach to say that even though (mn)^2 is positive, that it'd be less than kn+lm.

Thank you!

Let's start from here: (kn+lm)/mn < mn
In order to derive the inequality (kn+lm)/mn < 0 from the How about any quality, we have to ask ourselves "What operation are we performing on the original fraction?"
It appears as though we're simply subtracting mn from ONE side of the inequality, which is not a permissible move.

Your next step however is totally fine.
IF we were certain that (kn+lm)/mn < 0, then we can multiply both sides of the inequality by mn to get kn+lm > 0

However, as it stands, we can't logically take (kn+lm)/mn < mn and conclude that (kn+lm)/mn < 0

Does that help?

Thank you Brent