Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 29 Aug 2016, 00:28

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# What range of values of x will satisfy the inequality |2x +

Author Message
TAGS:

### Hide Tags

Intern
Joined: 05 Jan 2009
Posts: 1
Schools: ISB
Followers: 0

Kudos [?]: 22 [1] , given: 0

What range of values of x will satisfy the inequality |2x + [#permalink]

### Show Tags

25 May 2011, 07:04
1
KUDOS
21
This post was
BOOKMARKED
00:00

Difficulty:

55% (hard)

Question Stats:

65% (02:55) correct 35% (01:58) wrong based on 611 sessions

### HideShow timer Statistics

What range of values of x will satisfy the inequality |2x + 3| > |7x - 2|?

A. x < -1/9 or x > 5

B. -1 < x < 1/9

C. -1/9 < x < 1

D. -1/9 < x < 5

E. x < -1/9 or x > 1
[Reveal] Spoiler: OA

Last edited by Bunuel on 28 Apr 2014, 02:06, edited 2 times in total.
Renamed the topic, edited the question and the OA.
VP
Status: There is always something new !!
Affiliations: PMI,QAI Global,eXampleCG
Joined: 08 May 2009
Posts: 1353
Followers: 17

Kudos [?]: 217 [3] , given: 10

Re: What range of values of x will satisfy the inequality contd. [#permalink]

### Show Tags

25 May 2011, 07:26
3
KUDOS
1
This post was
BOOKMARKED
two methods here.

1. using the solutions here. x= -3/2 and x= 2/7 are solutions.
check for the regions

a. x<-3/2, b. -3/2<x<2/7 and c. x>2/7

values are a -(2x+3) > (7x-2) giving x <-1/9 hence not a solution.

b (2x+3) > (7x-2) giving x < 1 a solution.

c (2x+3) > -(7x-2) giving x>-1/9 hence the solution is -1/9 < x < 1

or

2. squaring both sides

(2x+3) ^2 > (7x-2)^2

gives 9x^2 -8x - 1 > 0 giving solution -1/9 < x < 1.

Hence C you can use whichever you are comfortable with.
_________________

Visit -- http://www.sustainable-sphere.com/
Promote Green Business,Sustainable Living and Green Earth !!

Intern
Joined: 05 Apr 2011
Posts: 10
Followers: 0

Kudos [?]: 3 [0], given: 0

Re: What range of values of x will satisfy the inequality contd. [#permalink]

### Show Tags

25 May 2011, 15:42
Amit,

Thank you, but could you please elaborate why in the option A

"a. x<-3/2,

values are a -(2x+3) > (7x-2) giving x <-1/9 hence not a solution."

the second part - (7x-2) is positive? If we get a number less than -1/9, -10 for instance would not the second part be negative as well?

Regards,
Steve.
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 11119
Followers: 511

Kudos [?]: 134 [0], given: 0

Re: What range of values of x will satisfy the inequality contd. [#permalink]

### Show Tags

27 Apr 2014, 14:15
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 34476
Followers: 6289

Kudos [?]: 79790 [7] , given: 10022

Re: What range of values of x will satisfy the inequality |2x + [#permalink]

### Show Tags

28 Apr 2014, 02:06
7
KUDOS
Expert's post
8
This post was
BOOKMARKED
ramzin wrote:
What range of values of x will satisfy the inequality |2x + 3| > |7x - 2|?

A. x < -1/9 or x > 5

B. -1 < x < 1/9

C. -1/9 < x < 1

D. -1/9 < x < 5

E. x < -1/9 or x > 1

Since both sides of the inequality are non-negative we can safely square:

$$4x^2+12x+9>49x^2-28x+4$$ --> $$45x^2-40x-5<0$$ --> $$9x^2-8x-1<0$$ --> $$(x+\frac{1}{9})(x-1)<0$$.

The "roots" are -1/9 and 1 (Solving Quadratic Inequalities: solving-quadratic-inequalities-graphic-approach-170528.html). "<" sign indicates that the solution must be between the roots: $$-\frac{1}{9}<x<1$$.

_________________
Intern
Joined: 05 Feb 2014
Posts: 48
Followers: 0

Kudos [?]: 13 [0], given: 49

Re: What range of values of x will satisfy the inequality |2x + [#permalink]

### Show Tags

28 Apr 2014, 04:27
Bunuel wrote:
ramzin wrote:
What range of values of x will satisfy the inequality |2x + 3| > |7x - 2|?

A. x < -1/9 or x > 5

B. -1 < x < 1/9

C. -1/9 < x < 1

D. -1/9 < x < 5

E. x < -1/9 or x > 1

Since both sides of the inequality are non-negative we can safely square:

$$4x^2+12x+9>49x^2-28x+4$$ --> $$45x^2-40x-5<0$$ --> $$9x^2-8x-1<0$$ --> $$(x+\frac{1}{9})(x-1)<0$$.

The "roots" are -1/9 and 1 (Solving Quadratic Inequalities: solving-quadratic-inequalities-graphic-approach-170528.html). "<" sign indicates that the solution must be between the roots: $$-\frac{1}{9}<x<1$$.

Hi Bunuel ,

Thanks for the answer. I did not understand how are the roots -1/9 and 1. I am getting the roots as 9 and 1. Can you please elaborate.
Math Expert
Joined: 02 Sep 2009
Posts: 34476
Followers: 6289

Kudos [?]: 79790 [0], given: 10022

Re: What range of values of x will satisfy the inequality |2x + [#permalink]

### Show Tags

28 Apr 2014, 04:38
gauravsoni wrote:
Bunuel wrote:
ramzin wrote:
What range of values of x will satisfy the inequality |2x + 3| > |7x - 2|?

A. x < -1/9 or x > 5

B. -1 < x < 1/9

C. -1/9 < x < 1

D. -1/9 < x < 5

E. x < -1/9 or x > 1

Since both sides of the inequality are non-negative we can safely square:

$$4x^2+12x+9>49x^2-28x+4$$ --> $$45x^2-40x-5<0$$ --> $$9x^2-8x-1<0$$ --> $$(x+\frac{1}{9})(x-1)<0$$.

The "roots" are -1/9 and 1 (Solving Quadratic Inequalities: solving-quadratic-inequalities-graphic-approach-170528.html). "<" sign indicates that the solution must be between the roots: $$-\frac{1}{9}<x<1$$.

Hi Bunuel ,

Thanks for the answer. I did not understand how are the roots -1/9 and 1. I am getting the roots as 9 and 1. Can you please elaborate.

$$(x+\frac{1}{9})(x-1)=0$$ --> $$x+\frac{1}{9}=0$$ or $$x-1=0$$ --> $$x=-\frac{1}{9}$$ or $$x=1$$.
_________________
Intern
Joined: 05 Feb 2014
Posts: 48
Followers: 0

Kudos [?]: 13 [0], given: 49

Re: What range of values of x will satisfy the inequality |2x + [#permalink]

### Show Tags

28 Apr 2014, 05:45
Bunuel wrote:
Bunuel wrote:
ramzin wrote:
What range of values of x will satisfy the inequality |2x + 3| > |7x - 2|?

A. x < -1/9 or x > 5

B. -1 < x < 1/9

C. -1/9 < x < 1

D. -1/9 < x < 5

E. x < -1/9 or x > 1

Since both sides of the inequality are non-negative we can safely square:

$$4x^2+12x+9>49x^2-28x+4$$ --> $$45x^2-40x-5<0$$ --> $$9x^2-8x-1<0$$ --> $$(x+\frac{1}{9})(x-1)<0$$.

The "roots" are -1/9 and 1 (Solving Quadratic Inequalities: solving-quadratic-inequalities-graphic-approach-170528.html). "<" sign indicates that the solution must be between the roots: $$-\frac{1}{9}<x<1$$.

Hi Bunuel ,

Thanks for the answer. I did not understand how are the roots -1/9 and 1. I am getting the roots as 9 and 1. Can you please elaborate.

$$(x+\frac{1}{9})(x-1)=0$$ --> $$x+\frac{1}{9}=0$$ or $$x-1=0$$ --> $$x=-\frac{1}{9}$$ or $$x=1$$.[/quote]

That part is fine , i'm solving the quadratic equation 9x^2-8x-1 < 0 as (x-9)(x+1) then getting x = 9 , x = -1. I also check out your link for solving quadratic equations in equalities but could relate it.
Math Expert
Joined: 02 Sep 2009
Posts: 34476
Followers: 6289

Kudos [?]: 79790 [0], given: 10022

Re: What range of values of x will satisfy the inequality |2x + [#permalink]

### Show Tags

28 Apr 2014, 05:52
gauravsoni wrote:
Bunuel wrote:
Bunuel wrote:
What range of values of x will satisfy the inequality |2x + 3| > |7x - 2|?

A. x < -1/9 or x > 5

B. -1 < x < 1/9

C. -1/9 < x < 1

D. -1/9 < x < 5

E. x < -1/9 or x > 1

Since both sides of the inequality are non-negative we can safely square:

$$4x^2+12x+9>49x^2-28x+4$$ --> $$45x^2-40x-5<0$$ --> $$9x^2-8x-1<0$$ --> $$(x+\frac{1}{9})(x-1)<0$$.

The "roots" are -1/9 and 1 (Solving Quadratic Inequalities: solving-quadratic-inequalities-graphic-approach-170528.html). "<" sign indicates that the solution must be between the roots: $$-\frac{1}{9}<x<1$$.

Hi Bunuel ,

Thanks for the answer. I did not understand how are the roots -1/9 and 1. I am getting the roots as 9 and 1. Can you please elaborate.

$$(x+\frac{1}{9})(x-1)=0$$ --> $$x+\frac{1}{9}=0$$ or $$x-1=0$$ --> $$x=-\frac{1}{9}$$ or $$x=1$$.

That part is fine , i'm solving the quadratic equation 9x^2-8x-1 < 0 as (x-9)(x+1) then getting x = 9 , x = -1. I also check out your link for solving quadratic equations in equalities but could relate it.[/quote]

(x-9)(x+1) is not a correct factoring of 9x^2-8x-1, it should be $$(x+\frac{1}{9})(x-1)$$ ($$(9x+1)(x-1)$$).

Hope this helps.
_________________
Intern
Joined: 09 Nov 2013
Posts: 11
Followers: 1

Kudos [?]: 9 [0], given: 17

Re: What range of values of x will satisfy the inequality |2x + [#permalink]

### Show Tags

29 Apr 2014, 01:49
amit2k9 wrote:
two methods here.

1. using the solutions here. x= -3/2 and x= 2/7 are solutions.
check for the regions

a. x<-3/2, b. -3/2<x<2/7 and c. x>2/7

values are a -(2x+3) > (7x-2) giving x <-1/9 hence not a solution.

b (2x+3) > (7x-2) giving x < 1 a solution.

c (2x+3) > -(7x-2) giving x>-1/9 hence the solution is -1/9 < x < 1

or

2. squaring both sides

(2x+3) ^2 > (7x-2)^2

gives 9x^2 -8x - 1 > 0 giving solution -1/9 < x < 1.

Hence C you can use whichever you are comfortable with.

Hi Bunuel,
Thanks for the solution! I get the squaring both sides approach, but it takes over 3 minutes for me to do it that way.
Is there a faster way to solve this?

I noticed Amit provided another method, but I'm not sure I understand the first approach completely. Could you explain why he used the "-" sign in the first and third parts (highlighted in blue above)?
Moderator
Joined: 25 Apr 2012
Posts: 728
Location: India
GPA: 3.21
Followers: 42

Kudos [?]: 615 [2] , given: 723

Re: What range of values of x will satisfy the inequality |2x + [#permalink]

### Show Tags

29 Apr 2014, 04:46
2
KUDOS
gauravsoni wrote:
Bunuel wrote:
ramzin wrote:
What range of values of x will satisfy the inequality |2x + 3| > |7x - 2|?

A. x < -1/9 or x > 5

B. -1 < x < 1/9

C. -1/9 < x < 1

D. -1/9 < x < 5

E. x < -1/9 or x > 1

Since both sides of the inequality are non-negative we can safely square:

$$4x^2+12x+9>49x^2-28x+4$$ --> $$45x^2-40x-5<0$$ --> $$9x^2-8x-1<0$$ --> $$(x+\frac{1}{9})(x-1)<0$$.

The "roots" are -1/9 and 1 (Solving Quadratic Inequalities: solving-quadratic-inequalities-graphic-approach-170528.html). "<" sign indicates that the solution must be between the roots: $$-\frac{1}{9}<x<1$$.

Hi Bunuel ,

Thanks for the answer. I did not understand how are the roots -1/9 and 1. I am getting the roots as 9 and 1. Can you please elaborate.

Hi Gauravsoni,

$$9x^2-8x-1<0$$

Can be factored 9x^2 -9x+x-1 <0

9x(x-1)+1 (x-1) <0 or (9x+1)(x-1)<0 or x=-1/9 or x=1
_________________

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

Manager
Joined: 20 Oct 2013
Posts: 66
Followers: 0

Kudos [?]: 2 [0], given: 27

Re: What range of values of x will satisfy the inequality |2x + [#permalink]

### Show Tags

07 May 2014, 10:41
hi bunnel

what do u mean when u say the following:

Quote:
Since both sides of the inequality are non-negative we can safely square:

4x^2+12x+9>49x^2-28x+4 --> 45x^2-40x-5<0 --> 9x^2-8x-1<0 --> (x+\frac{1}{9})(x-1)<0.

how how we solve inequality which is negative??
whenever we see absolute values on each side of the sign... should we use the squaring approach??
_________________

Hope to clear it this time!!
GMAT 1: 540
Preparing again

Math Expert
Joined: 02 Sep 2009
Posts: 34476
Followers: 6289

Kudos [?]: 79790 [1] , given: 10022

Re: What range of values of x will satisfy the inequality |2x + [#permalink]

### Show Tags

08 May 2014, 03:24
1
KUDOS
Expert's post
nandinigaur wrote:
hi bunnel

what do u mean when u say the following:

Quote:
Since both sides of the inequality are non-negative we can safely square:

4x^2+12x+9>49x^2-28x+4 --> 45x^2-40x-5<0 --> 9x^2-8x-1<0 --> (x+\frac{1}{9})(x-1)<0.

how how we solve inequality which is negative??
whenever we see absolute values on each side of the sign... should we use the squaring approach??

We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality)

As for your other questions: it depends on a question.
_________________
Senior Manager
Joined: 28 Apr 2014
Posts: 291
Followers: 1

Kudos [?]: 32 [0], given: 46

Re: What range of values of x will satisfy the inequality |2x + [#permalink]

### Show Tags

12 May 2014, 00:13
Bunuel wrote:
ramzin wrote:
What range of values of x will satisfy the inequality |2x + 3| > |7x - 2|?

A. x < -1/9 or x > 5

B. -1 < x < 1/9

C. -1/9 < x < 1

D. -1/9 < x < 5

E. x < -1/9 or x > 1

Since both sides of the inequality are non-negative we can safely square:

$$4x^2+12x+9>49x^2-28x+4$$ --> $$45x^2-40x-5<0$$ --> $$9x^2-8x-1<0$$ --> $$(x+\frac{1}{9})(x-1)<0$$.

The "roots" are -1/9 and 1 (Solving Quadratic Inequalities: solving-quadratic-inequalities-graphic-approach-170528.html). "<" sign indicates that the solution must be between the roots: $$-\frac{1}{9}<x<1$$.

If we put in the value x = 0 , the inequality is not being satisfied , even though it is within the range. Am I missing something here ?..

I tried to deduce the answers by substituting possible values from the various option ranges.
Math Expert
Joined: 02 Sep 2009
Posts: 34476
Followers: 6289

Kudos [?]: 79790 [0], given: 10022

Re: What range of values of x will satisfy the inequality |2x + [#permalink]

### Show Tags

12 May 2014, 00:18
himanshujovi wrote:
Bunuel wrote:
ramzin wrote:
What range of values of x will satisfy the inequality |2x + 3| > |7x - 2|?

A. x < -1/9 or x > 5

B. -1 < x < 1/9

C. -1/9 < x < 1

D. -1/9 < x < 5

E. x < -1/9 or x > 1

Since both sides of the inequality are non-negative we can safely square:

$$4x^2+12x+9>49x^2-28x+4$$ --> $$45x^2-40x-5<0$$ --> $$9x^2-8x-1<0$$ --> $$(x+\frac{1}{9})(x-1)<0$$.

The "roots" are -1/9 and 1 (Solving Quadratic Inequalities: solving-quadratic-inequalities-graphic-approach-170528.html). "<" sign indicates that the solution must be between the roots: $$-\frac{1}{9}<x<1$$.

If we put in the value x = 0 , the inequality is not being satisfied , even though it is within the range. Am I missing something here ?..

I tried to deduce the answers by substituting possible values from the various option ranges.

If x = 0, then |2x + 3| = 3 and |7x - 2| = 2 --> 3 > 2. So, 0 is a possible value of x.

Hope it helps.
_________________
Intern
Joined: 05 Feb 2014
Posts: 48
Followers: 0

Kudos [?]: 13 [0], given: 49

Re: What range of values of x will satisfy the inequality contd. [#permalink]

### Show Tags

18 Jun 2014, 06:15
amit2k9 wrote:
two methods here.

1. using the solutions here. x= -3/2 and x= 2/7 are solutions.
check for the regions

a. x<-3/2, b. -3/2<x<2/7 and c. x>2/7

values are a -(2x+3) > (7x-2) giving x <-1/9 hence not a solution.

b (2x+3) > (7x-2) giving x < 1 a solution.

c (2x+3) > -(7x-2) giving x>-1/9 hence the solution is -1/9 < x < 1

or

2. squaring both sides

(2x+3) ^2 > (7x-2)^2

gives 9x^2 -8x - 1 > 0 giving solution -1/9 < x < 1.

Hence C you can use whichever you are comfortable with.

Hi Amit, can you explain how to get the range -1/9 < x < 1 in the second method
the solutions are x< 1 and x < -1/9 (how does the sign reverse to the other side ?)
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 11119
Followers: 511

Kudos [?]: 134 [0], given: 0

Re: What range of values of x will satisfy the inequality |2x + [#permalink]

### Show Tags

23 Jul 2015, 14:22
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Manager
Joined: 24 May 2013
Posts: 86
Followers: 0

Kudos [?]: 14 [0], given: 99

Re: What range of values of x will satisfy the inequality |2x + [#permalink]

### Show Tags

28 Mar 2016, 08:03
What range of values of x will satisfy the inequality |2x + 3| > |7x - 2|?

Plotting the The halfs of the equation we get the diagram as shown...shaded portion shows the area of interest satisfyin the above inequality.
In upper half x=1 is the point of intersection i.e the maximum value of x. Only C satisfies.

A. x < -1/9 or x > 5

B. -1 < x < 1/9

C. -1/9 < x < 1

D. -1/9 < x < 5

E. x < -1/9 or x > 1
Attachments

l2x3l greater thn l7x2l.png [ 10.44 KiB | Viewed 2180 times ]

SVP
Joined: 17 Jul 2014
Posts: 1637
Location: United States
Schools: Stanford '19
GMAT 1: 550 Q39 V27
GMAT 2: 560 Q42 V26
GMAT 3: 560 Q43 V24
GMAT 4: 650 Q49 V30
GPA: 3.56
WE: General Management (Transportation)
Followers: 12

Kudos [?]: 194 [0], given: 109

Re: What range of values of x will satisfy the inequality |2x + [#permalink]

### Show Tags

10 Apr 2016, 19:27
ramzin wrote:
What range of values of x will satisfy the inequality |2x + 3| > |7x - 2|?

A. x < -1/9 or x > 5

B. -1 < x < 1/9

C. -1/9 < x < 1

D. -1/9 < x < 5

E. x < -1/9 or x > 1

I took the safest approach, plug in numbers :D

A. x=-2.
-4+3 = -1, |-1|=1. 7*-2 = -14 -2 = -16. |-16|=16 2 is not > than 16, so can't be A.
B. x= -1/8. -2/8+3 = 2.75. 7*-1/8 = -7/8 -2 = |-2.875| not true, so can't be B.
D. x=4. 8+3=11. 28-2=26. so not true, D is out.
E. same as in A. x=-2 doesn't work.

only C remains...
Intern
Joined: 23 Jan 2016
Posts: 39
Followers: 0

Kudos [?]: 1 [0], given: 297

Re: What range of values of x will satisfy the inequality |2x + [#permalink]

### Show Tags

24 Apr 2016, 21:38
Hi, I have some doubts.

In these kind of questions we generally get 4 results for x. In this question for instance, we get the following -

1. x<1
2. x>1
3. x<-1/9
4. x>-1/9

So should now the range be the inequality that restricts the the values within two limits? for example 1<x<-1/9 and cannot be the range right?

Thank you.
Re: What range of values of x will satisfy the inequality |2x +   [#permalink] 24 Apr 2016, 21:38
Similar topics Replies Last post
Similar
Topics:
5 How many integrals value of x satisfy the inequality 4 25 Feb 2016, 05:27
2 Which of the following values of x satisfy both inequalities 5 29 Nov 2012, 10:25
122 How many of the integers that satisfy the inequality (x+2)(x 32 09 Jun 2012, 00:00
56 For what range of values of x will the inequality 22 17 Feb 2011, 00:20
23 For what range of values of 'x' will the inequality 15x - 2/x > 1? 31 03 Sep 2010, 07:02
Display posts from previous: Sort by