Lesson: Inequalities - Part I

Comment on Inequalities - Part I

I solved it in another way and the result is slightly different.

5(1-2x)+7x > 4-x+9+2x

5-3x > 13+x
-2x > 8
x < -4

Is it right also? Why does this difference occur?
greenlight-admin's picture

Are you referring to a specific practice question or example?

That said, the third step in your solution is incorrect.

Take: 5 - 3x > 13 + x
Subtract x from both sides to get: 5 - 4x > 13
Subtract 5 from both sides to get: -4x > 8
Divide both sides by -4 to get: x < -2


Hello Brent,

For the following question, when i try to solve by isolating variables to one side, I end up having 1 (in Qty A) and 2y-x in (QTY B). 2y can be maximum when y=1/2. As 1-x will result in a value less than 1, I got result as A.

However, the response given by Sandy also seems valid (That the solution is D). Am I missing anything in my calculation ? Can you please help me ?


Thanks :)
greenlight-admin's picture

Question link: http://gre.myprepclub.com/forum/given-that-0-x-y-1806.html

Your statement that "2y can be maximum when y = 1/2" is incorrect.

All we know about y is that it's greater than both 0 and x, AND it's less than 1. So, y could be something like 0.9 or even 0.99999

Got my mistake. Thanks (y)

Brent, i am so confused with this one... please help me out...
2 ≤ |n| ≤ 5 how is n getting negative values?
greenlight-admin's picture

Hello Brent!
I had a doubt in this reinforcement question

If x, y, and z are positive numbers such that 3x < 2y < 4z, which of the following statements could be true?

Indicate all such statements.

A) x = y

B) y = z

C) y > z

D) x > z

For this question is it the case that if u find any such possible values for x y or z then u can just go ahead and mark the answer correct.
and what if u find a value which dosen't hold true for eg in choice D if we take x=4 and z=3 then we get that the inequality is equal and not 3x<4z .Still choice D is marked as a correct answer
I had this confusion .please help!
greenlight-admin's picture

This comes down to MUST be true versus COULD be true.

If the question asks us "What COULD be true?", then we need only find one instance where that thing is true. If we find an instance where a statement is not true, we can't eliminate that answer, since it doesn't rule out the possibility that the statement COULD be true.

If the question asks us "What MUST be true?", then we can ELIMINATE a a statement if it turns out to be false. However, we must be certain that the statement is ALWAYS true before we can select it.


That part was more of an ASIDE, since your solution does not really apply to the above concepts.

In your solution, you are REVERSING the question.
The question says IF 3x < 2y < 4z, THEN what could be true.
So, for example, IF 3x < 2y < 4z, THEN could it be true that x > z (answer choice D).

In your solution, you are asking "IF x > z, THEN could it be true that 3x < 2y < 4z?

So, in order to check each of the 4 statements, you must first find values of x, y and z that satisfy the given inequality (3x < 2y < 4z).

Does that help?


I understand the rules of inequalities, whereas adding or subtracting doesn't affect the inequality, but I'm confused about how you know what to subtract, divide and multiply.
For example, In 11 > 3 - 2x > -2 , how do you know to subtract -3, then divide by 2 and so on?
greenlight-admin's picture

The principles for solving inequalities are almost identical to solving equations.

In both cases, our goal is to ISOLATE the variable.

For example, let's say we have the equation: 3 - 2x = 21
Our first step might be to subtract 3 from both sides to get: -2x = 18
Then we can divide both sides by -2 to get: x = -9

The only difference between solving inequalities and solving equations, is that one little proviso that says:
If we divide or multiply both sides of an INEQUALITY by a NEGATIVE value, then we must REVERSE the direction of the inequality symbol. That's the only difference. So, just pretend you're solving an equation, and you'll be fine (as long as you remember the proviso above)

So, if we have: 11 > 3 - 2x > -2
We can subtract 3 from all three sides to get: 8 > -2x > -5
Then we can divide all three sides by -2 to get: -4 < x < 2.5

For more on ISOLATING the variable, watch: https://www.greenlighttestprep.com/module/gre-algebra-and-equation-solvi...


I solved it like this
Factor out 2 from (2x+4y) to get 2(x+2y) since we're told x+2y>8 2(x+2y could be 18 20,22 for values greater than 9,10,11
greenlight-admin's picture


I think that works :-)

What are you referring to when you write "for values greater than 9, 10, 11"?


Since x+2y>8 then 2(x+2y) is 2*some value greater than 8. These values could 9,10,11. If 9 quantity A becomes lesser,hence the answer D
greenlight-admin's picture

Aha! Yes, that works.


What's the difference between could be through and must be true
greenlight-admin's picture

If we're asked what COULD be true, all we need is to show ONE instance where the statement is true.

Consider this example:

If x is an integer, and k = 3x + 1, which of the following COULD be true?
I. k is a negative number

Let's test some values:
If x = 1, then k = 4 (not negative)
If x = 2, then k = 7 (not negative)
If x = 3, then k = 10 (not negative)
If x = -5, then k = -14 (NEGATIVE)

Since it is possible for k to be negative, statement I COULD be true.

If we're asked what MUST be true, it must be the case that the statement is ALWAYS true.

Consider this example:

If x is an integer, and k = 3x + 1, which of the following MUST be true?
I. k is a negative number

From our work above, we already know that k can be either positive or negative.
Since k is NOT always negative, statement I is FALSE.

Does that help?



Thank you for such an amazing series. Getting to the point, in the following link, why B and C aren't the answers?
greenlight-admin's picture

Thanks for the kind words!

Here's my solution: https://gre.myprepclub.com/forum/if-which-x-y-0-of-the-following-inequal...


Thank you. It makes sense now.

GIVEN: 0 < a < 1/b < 1
Since b is POSITIVE, we can safely multiply all parts of the inequality by b to get: 0 < ab < 1 < b
Hmmm, this new inequality doesn't have b and b² (which is needed to check the answer choices)
So, let's take 0 < ab < 1 < b and multiply all parts by b to get: 0 < ab² < b < b²
Let's check the answer choices to see which ones adhere to the inequality b < b²
(B) b > a > a² > b² NO GOOD. ELIMINATE
(C) b² > a > a² > b WORKS. KEEP
(E) b² > b > a > a² WORKS. KEEP

We're left with 2 answer choices:
(C) b² > a > a² > b
(E) b² > b > a > a²

Answer choice C says a > b and answer choice E says b > a
So, which is it?

Well, we know that a < 1/b
So, it COULD be the case that a = 1/3 and b = 2, which means b > a

I did not understand this last step. Please Elaborate
greenlight-admin's picture

Question link: https://gre.myprepclub.com/forum/if-0-a-1-then-which-of-the-following-mu...

The given information tells us: 0 < a < 1/b < 1
So, I focused on the part that says: a < 1/b

I looked for values of a and b that satisfy this inequality.
One pair of values is a = 1/3 and b = 2, because when I plug the values into the inequality, we get: 1/3 < 1/2 (which works!)

Since it's possible that a = 1/3 and b = 2, we can see that answer choice E must be correct, because it says that b > a

Does that help?



Why can't we directly multiply both sides by xy in this question since we already know that xy is negative?
greenlight-admin's picture

Link: https://gre.myprepclub.com/forum/xy-14399.html

Great idea! MUCH faster!!
Here's how that approach looks: https://gre.myprepclub.com/forum/xy-14399.html#p39245



Is factoring out the negative 1 the only way to get this right? I have tried so many methods and I always end up with B as bigger than A (-1 vs 1 or x vs 0). I probably would have never ever thought of factoring out the -1 first :/. Am I doomed if such a question were to show up?
greenlight-admin's picture

Question link: https://gre.myprepclub.com/forum/x-8146.html

I don't believe that it's entirely necessary to factor out -1 to solve this question.
For example, we could first recognize that, if x > 1, then quantity B must be POSITIVE, which means we can divide both quantities by (-x)/(1-x).

I think that approach will eventually yield the correct answer.

Is this the correct approach:

Multiply both sides by (-x)/(1-x)
--> A: (-x^2)/(-x^2+1) B: (x^2)/(x^2-2x+1)
--> A: (1)/(1+1) B: (1)/(1-2x+1) [for this step specifically I'm not sure if we're allowed to do it]
--> A: 1/2 B: (1)/(-2x+2)
--> So A is 1/2 and B will always be negative
Therefore A > B
greenlight-admin's picture

Sorry, KB23. I meant to say we could DIVIDE both sides by (-x)/(1-x).
I've edited my response above.

As for your solution, we cannot say that (-x^2)/(-x^2+1) = (1)/(1+1)
Nor can we say that (x^2)/(x^2-2x+1) = (1)/(1-2x+1)

Let's see how the solution might look if we DIVIDE both quantities by (-x)/(1-x)

QUANTITY A: x/(x+1)
QUANTITY B: (-x)/(1-x)

Divide both quantities by (-x)/(1-x) to get:
QUANTITY A: x/(x+1) ÷ (-x)/(1-x) = [x/(x+1)][(1-x)/(-x)] = (x-x²)/(-x+x²)

Factor and simplify Quantity A:
QUANTITY A: (x-x²)/(-x+x²) = x(1-x)/x(-1+x) = (1-x)/(-1+x)

Since x > 1, we know that (1-x) is always NEGATIVE, and (-1+x) is always POSITIVE
We we get:

Answer: B


Hi Brent ,
For https://gre.myprepclub.com/forum/n-is-an-integer-and-k-is-not-an-integer-0-k-n-k-15613.html

these are the conditions

so I have considered k=0.5;
thus k+2=0.5+2=2.5
So now n<2.5
here quantity b is k+1=>0.5+1=1.5

s n<k+2 shich is <2.5
n can be even 1.5 so I had choosen D

Is this a correct approach

greenlight-admin's picture

Question link: https://gre.myprepclub.com/forum/n-is-an-integer-and-k-is-not-an-integer...

I have a feeling your solution is valid.
I'm just having a hard time understanding what specific values of n you are using in your solution.
Can you elaborate?


I havent considered any specific n value. just considered the condition that n<k+2 so n values may be all the values that are less than k+2



HI Brent,

just want to double check whether it is possible to solve the aforementioned exercise by, instead of going your way, flicking everything to the right side of the inequality and determening that x1= -1 and x2= 5. Would we be able to solve through such a method as well?
I understand your way of doing it, but just want to have a complete overview if my way of thinking is faulty.
greenlight-admin's picture

Question link: https://gre.myprepclub.com/forum/if-7-x-2-4x-12-which-of-the-following-m...

That's a good idea, but the problem is that the resulting quadratic doesn't quite factor the way you think it does. Here's what I mean:

If we take: 7 < x² - 4x + 12
And subtract 7 from both sides, we get: 0 < x² - 4x + 5
At this point, it may seem like we can factor to get: 0 < (x - 5)(x + 1), however, if we expand and simplify (x - 5)(x + 1), we get x² - 4x - 5 (not x² - 4x + 5)



For this problem the way I did it as similar. (x+y)/n -(x+z)/n < 0 simplify and we get (y-z)/n < 0 ? just want to check if my logic is right?
greenlight-admin's picture

Question link: https://gre.myprepclub.com/forum/x-y-n-x-z-n-and-x-2-z-y-14623.html

That logic is perfect, Ravin. Nice work!

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