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If |x + 1 | > 2x - 1, which of the following represents the correct ra

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If |x + 1 | > 2x - 1, which of the following represents the correct ra  [#permalink]

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New post Updated on: 31 Oct 2018, 04:39
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If |x + 1 | > 2x - 1, which of the following represents the correct range of values of x ?

A. x < 0
B. x < 2
C. -2 < x < 0
D. -1 < x < 2
E. 0 < x < 2

Source: ExpertsGlobal

chetan2u,
Shouldn't we consider the overlapping region of (x<0) and (x<2), and the range would then be x < 0 ? Is the OA given wrong ?

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Originally posted by TaN1213 on 18 Aug 2018, 10:23.
Last edited by Bunuel on 31 Oct 2018, 04:39, edited 2 times in total.
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Re: If |x + 1 | > 2x - 1, which of the following represents the correct ra  [#permalink]

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New post Updated on: 18 Aug 2018, 12:28
3
1
TaN1213 wrote:
if |x + 1 | > 2x - 1, which of the following represents the correct range of values of x ?

A. x < 0
B. x < 2
C. -2 < x < 0
D. -1 < x < 2
E. 0 < x < 2

Source: ExpertsGlobal

Shouldn't we consider the overlapping region of (x<0) and (x<2), and the range would then be x < 0 ? Is the OA given wrong ?


Given, |x + 1 | > 2x - 1
Or, x+1 > 2x-1 (|x+1|=x+1,when \((x+1)\geq{0}\) or, \(x\geq{-1}\))
Or, 2x-1< x+1
Or, x< 2
So, \(x\ge \:-1\quad \mathrm{and}\quad \:x<2\)-------------(1)

Given, |x + 1 | > 2x - 1
Or, -(x+1) > 2x-1 (|x+1|=-(x+1), when (x+1) < 0 Or, x< -1)
Or, 2x-1< -(x+1)
Or, 2x-1< -x-1
Or, 3x < 0
Or, x < 0
So, \(x<-1\quad \mathrm{and}\quad \:x<0\)----------------(2)

Combining (1) & (2),

\(\left(x<-1\quad \mathrm{and}\quad \:x<0\right)\quad \mathrm{or}\quad \left(x\ge \:-1\quad \mathrm{and}\quad \:x<2\right)\)
So, \(x < 2\)

Ans. (B)
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Originally posted by PKN on 18 Aug 2018, 10:35.
Last edited by PKN on 18 Aug 2018, 12:28, edited 1 time in total.
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Re: If |x + 1 | > 2x - 1, which of the following represents the correct ra  [#permalink]

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New post 18 Aug 2018, 11:36
1
PKN wrote:
TaN1213 wrote:
if |x + 1 | > 2x - 1, which of the following represents the correct range of values of x ?

A. x < 0
B. x < 2
C. -2 < x < 0
D. -1 < x < 2
E. 0 < x < 2

Source: ExpertsGlobal

Shouldn't we consider the overlapping region of (x<0) and (x<2), and the range would then be x < 0 ? Is the OA given wrong ?


Given, |x + 1 | > 2x - 1
Or, x+1 > 2x-1 (|x+1|=x+1,when (x+1) > 0)
Or, 2x-1< x+1
Or, x< 2-------------(1)

Given, |x + 1 | > 2x - 1
Or, -(x+1) > 2x-1 (|x+1|=-(x+1), when (x+1) < 0)
Or, 2x-1< -(x+1)
Or, 2x-1< -x-1
Or, 3x < 0
Or, x < 0---------------(2)

From (1) & (2), x < 2

Ans. (B)

Would you please help me understand why aren't we taking the overlap here and choose x<0 ?
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Re: If |x + 1 | > 2x - 1, which of the following represents the correct ra  [#permalink]

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New post 18 Aug 2018, 20:35
1
TaN1213 wrote:
if |x + 1 | > 2x - 1, which of the following represents the correct range of values of x ?

A. x < 0
B. x < 2
C. -2 < x < 0
D. -1 < x < 2
E. 0 < x < 2

Source: ExpertsGlobal

chetan2u,
Shouldn't we consider the overlapping region of (x<0) and (x<2), and the range would then be x < 0 ? Is the OA given wrong ?


Hi...
As correctly done , two critical points..
1) x\(\leq{-1}\)
-(x+1)>2x-1........x+1<1-2x......X<0
So all values below and equal to -1 satisfy the equation
2) x>-1
x+1>2x-1.......x<2
So all values between -1 and 2 satisfy the equation..

Combined..
All values upto -1 inclusive and all values between -1 and 2
So it turns out all values till 2
And that is why both combined give us x<2
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Re: If |x + 1 | > 2x - 1, which of the following represents the correct ra  [#permalink]

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New post 18 Aug 2018, 21:11
1
chetan2u wrote:
TaN1213 wrote:
if |x + 1 | > 2x - 1, which of the following represents the correct range of values of x ?

A. x < 0
B. x < 2
C. -2 < x < 0
D. -1 < x < 2
E. 0 < x < 2

Source: ExpertsGlobal

chetan2u,
Shouldn't we consider the overlapping region of (x<0) and (x<2), and the range would then be x < 0 ? Is the OA given wrong ?


Hi...
As correctly done , two critical points..
1) x\(\leq{-1}\)
-(x+1)>2x-1........x+1<1-2x......X<0
So all values below and equal to -1 satisfy the equation
2) x>-1
x+1>2x-1.......x<2
So all values between -1 and 2 satisfy the equation..

Combined..
All values upto -1 inclusive and all values between -1 and 2
So it turns out all values till 2
And that is why both combined give us x<2


Hi chetan2u

Whats wrong with this approach.Please explain.
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Re: If |x + 1 | > 2x - 1, which of the following represents the correct ra  [#permalink]

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New post 18 Aug 2018, 21:31
2
gmat1393 wrote:
chetan2u wrote:
TaN1213 wrote:
if |x + 1 | > 2x - 1, which of the following represents the correct range of values of x ?

A. x < 0
B. x < 2
C. -2 < x < 0
D. -1 < x < 2
E. 0 < x < 2

Source: ExpertsGlobal

chetan2u,
Shouldn't we consider the overlapping region of (x<0) and (x<2), and the range would then be x < 0 ? Is the OA given wrong ?


Hi...
As correctly done , two critical points..
1) x\(\leq{-1}\)
-(x+1)>2x-1........x+1<1-2x......X<0
So all values below and equal to -1 satisfy the equation
2) x>-1
x+1>2x-1.......x<2
So all values between -1 and 2 satisfy the equation..

Combined..
All values upto -1 inclusive and all values between -1 and 2
So it turns out all values till 2
And that is why both combined give us x<2


Hi chetan2u

Whats wrong with this approach.Please explain.


Hi..

You square both sides ONLY when you are sure both sides are positive...
That is the reason you are getting your values when X is positive.
But what about x as negative, 2x-1 will be negative too at those values.

So REMEMBER, do not square when both sides are not 100% to be positive
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Re: If |x + 1 | > 2x - 1, which of the following represents the correct ra  [#permalink]

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New post 18 Aug 2018, 21:34
Hi chetan2u.

I consistently get these mod questions and inequalities questions wrong. Could you please guide me as to how to improve and practice(i dont want to practice QUESTION PACK questions for these topics)

Posted from my mobile device
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Re: If |x + 1 | > 2x - 1, which of the following represents the correct ra  [#permalink]

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New post 28 Oct 2018, 02:52
PKN wrote:
TaN1213 wrote:
if |x + 1 | > 2x - 1, which of the following represents the correct range of values of x ?

A. x < 0
B. x < 2
C. -2 < x < 0
D. -1 < x < 2
E. 0 < x < 2

Source: ExpertsGlobal

Shouldn't we consider the overlapping region of (x<0) and (x<2), and the range would then be x < 0 ? Is the OA given wrong ?


Given, |x + 1 | > 2x - 1
Or, x+1 > 2x-1 (|x+1|=x+1,when \((x+1)\geq{0}\) or, \(x\geq{-1}\))
Or, 2x-1< x+1
Or, x< 2
So, \(x\ge \:-1\quad \mathrm{and}\quad \:x<2\)-------------(1)

Given, |x + 1 | > 2x - 1
Or, -(x+1) > 2x-1 (|x+1|=-(x+1), when (x+1) < 0 Or, x< -1)
Or, 2x-1< -(x+1)
Or, 2x-1< -x-1
Or, 3x < 0
Or, x < 0
So, \(x<-1\quad \mathrm{and}\quad \:x<0\)----------------(2)

Combining (1) & (2),

\(\left(x<-1\quad \mathrm{and}\quad \:x<0\right)\quad \mathrm{or}\quad \left(x\ge \:-1\quad \mathrm{and}\quad \:x<2\right)\)
So, \(x < 2\)

Ans. (B)




Hey PKN,

can you pls explain how after this x < 0 you concluded that \(x<-1\) x could be decimal as well -0.5 :)


thanks!
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Re: If |x + 1 | > 2x - 1, which of the following represents the correct ra  [#permalink]

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New post Updated on: 30 Oct 2018, 11:55
Hi chetan2u pushpitkc

can you please explain why the correct answer is not x<0 :?

see attached my solution, as far as I know when x<0 and x <2 we need to pick the common portion ( I highlighted it) as correct answer. Isn`t it so ? or are there are some exceptions of which I am not aware

so I chose A, can you explain what`s wrong with my reasoning ?

thanks!

Dave

p.s.bb why i am getting an italic font ? i ddnt use any font formatting tools


UPDATE
:)

thanks to chetan2u and PKN for taking time to explain but I still dont get WHY OVERLAPPING PORTION of my solution above is not correct ? :)

Have a look at the video below (first 5 minutes are enough :)), this Indian guy clearly explains that overlapping portions are answers to such questions



generis, pushpitkc Bunuel

or may be you can explain Probus, ganand, VeritasKarishma, GMATPrepNow :)
Attachments

Absolute Values1.png
Absolute Values1.png [ 37.4 KiB | Viewed 2957 times ]


Originally posted by dave13 on 28 Oct 2018, 03:21.
Last edited by dave13 on 30 Oct 2018, 11:55, edited 3 times in total.
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Re: If |x + 1 | > 2x - 1, which of the following represents the correct ra  [#permalink]

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New post 28 Oct 2018, 06:02
1
dave13 wrote:
Hi chetan2u pushpitkc

can you please explain why the correct answer is not x<0 :?

see attached my solution, as far as I know when x<0 and x <2 we need to pick the common portion ( I highlighted it) as correct answer. Isn`t it so ? or are there are some exceptions of which I am not aware

so I chose A, can you explain what`s wrong with my reasoning ?

thanks!

Dave

p.s.bb why i am getting an italic font ? i ddnt use any font formatting tools



Hi..
Substitute x as 1 and you will get the equation to be true so x as 1 is surely a value..
Now where you are going wrong..
Refer to my post about critical points I specific to this question or the post attached in the signature for a detailed explanation.

Now |x+1| can have two critical points as shown in my post above..
1) x<=-1.. here we get x<0 as solution...
So this means x<=-1 is correct as solution is x<0 so what we initially assume as x<=-1 is correct
3) x>-1 gives x<2 as solution
This means the values from -1 to 2 are correct as we had assumed that x>-1 at the start
When you combined this two you het x<=-1 and -1<x<2
Combined it is x<2
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Re: If |x + 1 | > 2x - 1, which of the following represents the correct ra  [#permalink]

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New post 28 Oct 2018, 09:15
1
Hi dave13,
Given, |x + 1 | > 2x - 1
Or, -(x+1) > 2x-1 (|x+1|=-(x+1), when (x+1) < 0 Or, x< -1)
Or, 2x-1< -(x+1)
Or, 2x-1< -x-1
Or, 3x < 0
Or, x < 0
Therefore,
a) x+1<0 Or x<-1
b) x<0
So, \(x<-1\quad \mathrm{and}\quad \:x<0\)

dave13 wrote:
PKN wrote:
TaN1213 wrote:
if |x + 1 | > 2x - 1, which of the following represents the correct range of values of x ?

A. x < 0
B. x < 2
C. -2 < x < 0
D. -1 < x < 2
E. 0 < x < 2

Source: ExpertsGlobal

Shouldn't we consider the overlapping region of (x<0) and (x<2), and the range would then be x < 0 ? Is the OA given wrong ?


Given, |x + 1 | > 2x - 1
Or, x+1 > 2x-1 (|x+1|=x+1,when \((x+1)\geq{0}\) or, \(x\geq{-1}\))
Or, 2x-1< x+1
Or, x< 2
So, \(x\ge \:-1\quad \mathrm{and}\quad \:x<2\)-------------(1)

Given, |x + 1 | > 2x - 1
Or, -(x+1) > 2x-1 [highlight](|x+1|=-(x+1), when (x+1) < 0 Or, x< -1)

Or, 2x-1< -(x+1)
Or, 2x-1< -x-1
Or, 3x < 0
Or, x < 0[/highlight]
So, \(x<-1\quad \mathrm{and}\quad \:x<0\)----------------(2)

Combining (1) & (2),

\(\left(x<-1\quad \mathrm{and}\quad \:x<0\right)\quad \mathrm{or}\quad \left(x\ge \:-1\quad \mathrm{and}\quad \:x<2\right)\)
So, \(x < 2\)

Ans. (B)




Hey PKN,

can you pls explain how after this x < 0 you concluded that \(x<-1\) x could be decimal as well -0.5 :)


thanks!

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Re: If |x + 1 | > 2x - 1, which of the following represents the correct ra  [#permalink]

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New post 30 Oct 2018, 18:01
1
Top Contributor
TaN1213 wrote:
if |x + 1 | > 2x - 1, which of the following represents the correct range of values of x ?

A. x < 0
B. x < 2
C. -2 < x < 0
D. -1 < x < 2
E. 0 < x < 2


I find that the quickest solutions to this kind of question involve testing the answer choices

Scan the answer choices
Notice that some answer choices say that x = 1 is a solution and some say x = 1 is NOT a solution.
So, let's test x = 1
Plug it into the original inequality to get: |1 + 1 | > 2(1) - 1
Simplify to get: 2 > 1
Perfect!
So, x = 1 IS a solution to the inequality.

Since answer choices A and C do NOT include x = 1 as a solution, we can ELIMINATE them.

Now scan the remaining answer choices (B, D and E)
Some answer choices say that x = -1 is a solution and some say x = -1 is NOT a solution.
So, let's test x = -1
Plug it into the original inequality to get: |(-1) + 1 | > 2(-1) - 1
Simplify to get: 0 > -3
Perfect!
So, x = -1 IS a solution to the inequality.

Since answer choices D and E do NOT include x = -1 as a solution, we can ELIMINATE them.

We're left with B

Answer: B

Cheers,
Brent
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Re: If |x + 1 | > 2x - 1, which of the following represents the correct ra  [#permalink]

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New post 31 Oct 2018, 03:06
Hello Brent :) thank you for your alternative approach but all I wanted is to understand why in the example below we take an overlap as a solution and in the question above we DO NOT TAKE overlap as solution.

EXAMPLE:

X^2+5x+6>0
(x+2)(x+3)>0

Now we have two options either both numbers are positive or both numbers are negative.


Option 1. (both are positive)

x+2>0 ---> x>-2

x+3>0 ---> x>-3

now look here attentively so we have x>-2 and x>-3 (now as per the logic of above mentioned solutions the interval should be x>-3 because it includes -2 as well) Excusez-moi monsieur, :) but it is not correct, the correct interval is x>-2 i.e. the overlapping interval !


See attached.
Attachment:
Interval One .png
Interval One .png [ 19.66 KiB | Viewed 2559 times ]



Option 2. (both are negative)

x+2<0 ---> x<0

x+3<0 ---> x<-3

now look here carefully so we have x<0 and x<-3 (now as per the logic of above mentioned solutions the interval should be x<-2 because it includes -3 as well) Excusez-moi monsieur, but it is not correct, the correct interval is x<-3 i.e. the overlapping interval !


See attached.
Attachment:
Interval Two.png
Interval Two.png [ 20.51 KiB | Viewed 2563 times ]


Thank you! :)
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Re: If |x + 1 | > 2x - 1, which of the following represents the correct ra  [#permalink]

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New post 31 Oct 2018, 04:33
TaN1213 wrote:
if |x + 1 | > 2x - 1, which of the following represents the correct range of values of x ?

A. x < 0
B. x < 2
C. -2 < x < 0
D. -1 < x < 2
E. 0 < x < 2

Source: ExpertsGlobal

chetan2u,
Shouldn't we consider the overlapping region of (x<0) and (x<2), and the range would then be x < 0 ? Is the OA given wrong ?

This particular question you can solve it either by inequality or by putting smart numbers.
just look at the options
put X=0
given equation is valid. It means options A, C and E are out.
then it is between B and D
put X= -1
given equation is valid. D is out.
answer B:)
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If |x + 1 | > 2x - 1, which of the following represents the correct ra  [#permalink]

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New post Updated on: 01 Nov 2018, 01:41
hello Bunuel, pushpitkc Gladiator59, chetan2u ,
generis,
in other words friendly modhumanbeings :) its me dave13, designer of hamburgers & hotdogs, :lol: as you see my profession is far from math, so can you please provide me with ps examples/links/ps questions of absolute values with an overlap on number line :) Thank you! :)

Originally posted by dave13 on 01 Nov 2018, 00:59.
Last edited by dave13 on 01 Nov 2018, 01:41, edited 2 times in total.
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Re: If |x + 1 | > 2x - 1, which of the following represents the correct ra  [#permalink]

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New post 01 Nov 2018, 01:22
1
Hello dave13!

Hope you are doing great! Designing hot-dogs sounds fun, is it as much fun as eating one? :-P

For absolute value material - there are tons of links on gmatclub. I personally read through veritas prep blog on absolute values and also Bunuel's post also walker's post here which is a part of the GMAT club math book.

In addition you might want to check out the youtube channel of khan academy who are doing a great job providing free education through out the world. Here is an absolute values section that has some exercises as well. ( Be careful though as not all of it is part of GMAT but if you are struggling with graphical approach I would highly recommend it)

For PS/DS question kindly use the GMAT Club question bank with the appropriate filters.

Feel free to tag me in more specific questions.

Best,
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Re: If |x + 1 | > 2x - 1, which of the following represents the correct ra  [#permalink]

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New post 06 Nov 2018, 14:39
Hi All,

We're told that |X + 1 | > 2X - 1. We're asked which of the following represents the correct range of values of X. This question can be solved in a couple of different ways, including by TESTing VALUES.

Let's start with the easiest number possible: 0

Will X=0 fit the given inequality?.... |0 + 1| > 2(0) - 1 ..... 1 > -1. Yes, this 'fits', so X=0 IS a possible solution.
Eliminate Answers A, C and E.

Between the remaining 2 answers (Answer B and Answer D), some values of X 'overlap' and some don't, so we should focus on the values that DON'T overlap...

Will X= -2 fit the given inequality?.... |0 + -2| > 2(-2) - 1 ..... 2 > -5. Yes, this 'fits', so X= -2 IS a possible solution.
Eliminate Answer D.

Final Answer: B

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Re: If |x + 1 | > 2x - 1, which of the following represents the correct ra  [#permalink]

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New post 16 Dec 2018, 09:10
If x = -6

(-6+1) = -5

2(-6)-1 = -13

5 is not greater than 13.. how is the solution x < 2? LHS is not greater than RHS.. someone plz explain.

x = 3 does not work --> (3+1) < 2(3)-1
x = 2 does not work --> (2+1) = 4-1
x = 1 works --> (1+1) > 1
x = 0 works --> (0+1) > -1
x = -1 does not work -> (-1+1) < -3

Don't get at all.
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Re: If |x + 1 | > 2x - 1, which of the following represents the correct ra  [#permalink]

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New post 16 Dec 2018, 14:43
Hi PierTotti17,

The question asks us to find the range of values for the inequality |X + 1| > 2X - 1. You can approach this work in a couple of different ways, including TESTing VALUES (meaning that you can eliminate Answer choices based on values for X that either "fit" or "don't fit" the given inequality.

You mention X = -6 in your post...

IF... X = -6, then we have...

|-6 + 1| = |-5| = +5
(2)(-6) - 1 = -12 - 1 = -13

+5 is greater than -13, so X = -6 IS a solution to the given inequality. Thus, the correct answer has to include X = -6 in its range. With this one TEST, you can eliminate Answers C, D and E.

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If |x + 1 | > 2x - 1, which of the following represents the correct ra  [#permalink]

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New post 18 Dec 2018, 07:48
TaN1213 wrote:
If |x + 1 | > 2x - 1, which of the following represents the correct range of values of x ?

A. x < 0
B. x < 2
C. -2 < x < 0
D. -1 < x < 2
E. 0 < x < 2

Source: ExpertsGlobal



Gladiator59 can you please check my solution below, and reveal any (even slightest) flaws :) Thank you !

So we have, |x + 1 | > 2x - 1

Set condition when x is positive
|x + 1 | ≥ 0
x≥-1

Check if equation below satisfies given condition:
|x + 1 | > 2x - 1
x+1>2x-1
x<2

So we know x<2 with conditions that it X is positive and x≥-1 ( hence x could be 0.1, 0.2 , etc inclusive 1.99 )


Set condition when x is negative
|x + 1 | < 0
x+1<0
x <-1

Check if equation below satisfies given condition:
-|x + 1 | > 2x - 1
-x-1>2x-1
x <0

So we know x < 0 with the condition X is negative and x <-1 ( hence x could be -1.1, -1.2, -3, -100 etc )


Now combing both conditions

So we know x < 2 with condition, X is positive and x ≥-1 ( hence x could be 0.1, 0.2 , etc inclusive 1.99 )

So we know x < 0 with condition, X is negative and x <-1 ( hence x could be -1.1, -1.2, -3, -100 etc )


On combining X could be 0.1, 0.2 , etc inclusive 1.99 and X could be -1.1, -1.2, -3, -100 etc


lets analyze answer choices

A. x < 0 (INCORRECT, X COULD BE x could be 0.1, 0.2 , etc inclusive 1.99 )

B. x < 2 ( CORRECT X could be 0.1, 0.2 , etc inclusive 1.99 and X could be -1.1, -1.2 , -3, -100 etc

C. -2 < x < 0 ( INCORRECT X could be -1.1, -1.2 , -3 , -100, etc )

D. -1 < x < 2 (INCORRECT X could be -1.1, -1.2 , -3 , -100, etc )

E. 0 < x < 2 ( INCORRECT X could be -1.1, -1.2 , -3 , -100, etc )
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If |x + 1 | > 2x - 1, which of the following represents the correct ra   [#permalink] 18 Dec 2018, 07:48

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