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

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

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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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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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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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

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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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: [spoiler]B[/spoiler]

GMAT assassins aren't born, they're made,
Rich

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.

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.

GMAT assassins aren't born, they're made,
Rich

Bunuel chetan2u KarishmaB ScottTargetTestPrep IanStewart

could you please help me understand why we are not taking only the overlap of the solutions of x ie. answer to be only x<0 but we take the overlap in this https://gmatclub.com/forum/is-x-217529.html#p2293754 question for statement I?

Thank you in advance!

Hi

In a DS question, if two statements give different ranges, the overlap is the answer as each statement is independent of others.

But here, it is PS and the same equation is giving two different ranges, and combination of both these that includes all points will be the answer.

So if, for example, this question would have x<0 as one range and other as x<2 as the other, the answer would be x<2, that is range that includes all points.
But, if a DS gives these two ranges as ranges from two statements, the answer will be x<0, that is range that is common to two statements.
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I suppose it depends how you solve the question. We have this inequality:

|x + 1| > 2x - 1

which is clearly true for any negative number, because the left side will be positive and the right side will be negative. So we know the answer can only be A or B, and you can just test x = 1 to see that B must be right.

If you do solve these kinds of questions algebraically (something that can almost always be avoided on the GMAT -- there are a few other things you could do) then things can get a bit annoying. I'll explain with a simpler inequality:

|z| < 5

You don't need to use algebra here, but supposing you wanted to, the standard technique is to divide the problem into two cases:

• if the thing inside the absolute value is positive, then the absolute value won't do anything. So here, if z > 0, then |z| = z, and our inequality becomes z < 5. Notice though that we assumed from the outset that z > 0, so our work is only valid when z > 0. So here we must take the 'overlap' of our assumption and our solution, and we find 0 < z < 5 is one set of solutions to the inequality

• if the thing inside the absolute value is negative, then the absolute value will flip its sign. So if z < 0, then |z| = -z, and our inequality becomes -z < 5, or z > -5. Again, we have to combine our solution with our assumption, so a second set of solutions is -5 < z < 0.

You can throw z = 0 into either case, and it won't matter which (z = 0 is also a solution here). So we now have two separate sets of legitimate solutions to our inequality, and since they all work, we now want to combine them: -5 < z < 5 is the complete set of solutions.

So when you analyze each individual case, you must take the overlap of your assumption and your solution, but since each case will, most of the time, give you some valid solutions to the inequality, when you combine the cases you use all of the solutions you found.

ColumbiaBaby

Further, check out this post: https://anaprep.com/algebra-the-why-beh ... questions/
Even after doing many absolute value questions, often we miss this point. When we break the number line into two ( x >= 0 and x < 0) we need to combine the solutions obtained in both cases because we are considering only half the values in each case.
So if we get that when x >= 0, x can be 0, 1, 2 or 3 and when x < 0, x can be -1 or -2, we ar essentially saying that x can take integer values from -2 to 3.

Essentially, the difference between the two questions is the difference between the way we solve an inequality of the form |y| < c and an inequality of the form |y| > c. The former is only satisfied when y is between -c and c (assuming c is positive), and the latter is satisfied when y is greater than c or when y is less than -c. As you noticed, the solution set of |y| < c is actually "0 ≤ y < c or -c < y ≤ 0", but this is identical to -c < y < c.
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Given that |x + 1 | > 2x - 1 and we need to find the correct range of values of x

Let's solve the problem using two methods

Method 1: Substitution

We will values in each option choice and plug in the question and check if it satisfies the question or not. ( Idea is to take such values which can prove the question wrong)

A. x < 0

Lets take x = -1 (which falls in this range of x < 0) and substitute in the equation |x + 1 | > 2x - 1
=> |-1 + 1 | > 2*-1 - 1
=> | 0| > -3
=> 0 > -3 which is TRUE, but let's see if we can find a bigger range

B. x < 2

Lets take x = 1.9 (which falls in this range of x < 2) and substitute in the equation |x + 1 | > 2x - 1
=> |1.9 + 1 | > 2*1.9 - 1
=> | 2.9 | > 2.8
=> 0 > -3 which is TRUE.
Now, this range is bigger than option A, so we can ignore option A now.

C. -2 < x < 0

This is a subset of Option B so we don't need to check this one

D. -1 < x < 2

This is a subset of Option B so we don't need to check this one

E. 0 < x < 2

This is a subset of Option B so we don't need to check this one

So, Answer will be B

Method 2: Algebra

Now, we know that |A| > B can be opened as (Watch this video to know about the Basics of Absolute Value)
A > B for A ≥ 0 and
-A > B for A < 0

=> |x + 1 | > 2x - 1 can be written as

Case 1: x + 1 ≥ 0 or x ≥ -1
=> x + 1 > 2x - 1
=> 2x - x < 1 + 1
=> x < 2
And the condition was x ≥ -1 and we got the answer as x < 2
=> the solution will be the range common in both of them
=> -1 ≤ x < 2



Case 2: x + 1 < 0 or x < -1
=> -(x + 1) > 2x - 1
=> 2x + x < 1 - 1
=> 3x < 0
=> x < 0
And the condition was x < -1 and we got the answer as x < 0
=> the solution will be the range common in both of them
=> x < -1



Combining both the cases we get the answer as -1 ≤ x < 2 and x < -1
=> x < 2

So, Answer will be B
Hope it helps!

Watch the following video to learn How to Solve Absolute Value Problems

[youtu-be]https://www.youtube.com/watch?v=rBWfl4AmQDI[/you-tube]
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