Author
Message
TAGS:

Hide Tags
Manager

Joined: 02 Dec 2012

Posts: 178

Followers: 5

Kudos [? ]:
2430
[0 ] , given: 0

If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink ]

Show Tags
27 Dec 2012, 05:45

23

This post was BOOKMARKED

Question Stats:

68% (02:24) correct

32% (02:06) wrong

based on 1081 sessions

Hide Show timer Statistics
\((\frac{x+1}{x-1})^2\)

If x#0 and x#1, and if x is replaced by 1/x everywhere in the expression above, then the resulting expression is equivalent to

A. \((\frac{x+1}{x-1})^2\)

B. \((\frac{x-1}{x+1})^2\)

C. \(\frac{x^2+1}{1-x^2}\)

D. \(\frac{x^2-1}{x^2+1}\)

E. \(-(\frac{x-1}{x+1})^2\)

Official Answer and Stats are available only to registered users.

Register /

Login .

Math Expert

Joined: 02 Sep 2009

Posts: 37080

Followers: 7242

Kudos [? ]:
96313
[5 ]
, given: 10733

Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink ]

Show Tags
27 Dec 2012, 05:56
5

This post received KUDOS

Expert's post

7

This post was BOOKMARKED

Manager

Joined: 04 Dec 2011

Posts: 81

Followers: 0

Kudos [? ]:
25
[5 ]
, given: 13

Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink ]

Show Tags
02 May 2013, 14:38
5

This post received KUDOS

Bunuel wrote:

[b]\((\frac{x+1}{x-1})^2\) \((\frac{\frac{1}{x}+1}{\frac{1}{x}-1})^2=(\frac{\frac{1+x}{x}}{\frac{1-x}{x}})^2=(\frac{1+x}{1-x})^2=(\frac{x+1}{x-1})^2\) .

How did 1-X in the denominator become X-1 in last step?

_________________

Life is very similar to a boxing ring. Defeat is not final when you fall down… It is final when you refuse to get up and fight back! 1 Kudos = 1 thanks Nikhil

Intern

Joined: 23 Apr 2013

Posts: 22

Followers: 0

Kudos [? ]:
19
[6 ]
, given: 1

Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink ]

Show Tags
02 May 2013, 20:57
6

This post received KUDOS

2

This post was BOOKMARKED

nikhil007 wrote:

Bunuel wrote:

[b]\((\frac{x+1}{x-1})^2\) \((\frac{\frac{1}{x}+1}{\frac{1}{x}-1})^2=(\frac{\frac{1+x}{x}}{\frac{1-x}{x}})^2=(\frac{1+x}{1-x})^2=(\frac{x+1}{x-1})^2\) .

How did 1-X in the denominator become X-1 in last step?

It's not (1-x) that became (x-1).

\((1-x)^2\) is simply rewritten as \((x-1)^2\).

Both \((1-x)^2\) and \((x-1)^2\) are essentially the same

Verbal Forum Moderator

Joined: 10 Oct 2012

Posts: 630

Followers: 82

Kudos [? ]:
1136
[3 ]
, given: 136

Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink ]

Show Tags
02 May 2013, 21:03
3

This post received KUDOS

1

This post was BOOKMARKED

nikhil007 wrote:

Bunuel wrote:

[b]\((\frac{x+1}{x-1})^2\) \((\frac{\frac{1}{x}+1}{\frac{1}{x}-1})^2=(\frac{\frac{1+x}{x}}{\frac{1-x}{x}})^2=(\frac{1+x}{1-x})^2=(\frac{x+1}{x-1})^2\) .

How did 1-X in the denominator become X-1 in last step?

\((1-x)^2 = (x-1)^2\). For example, \((1-4)^2 = (4-1)^2 = 9\).

The negative sign inside the bracket gets taken care of because of the square.

_________________

All that is equal and not-Deep Dive In-equality Hit and Trial for Integral Solutions

Manager

Joined: 29 Mar 2010

Posts: 141

Location: United States

Concentration: Finance, International Business

GMAT 1 : 590 Q28 V38

GPA: 2.54

WE: Accounting (Hospitality and Tourism)

Followers: 1

Kudos [? ]:
117
[0 ] , given: 16

Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink ]

Show Tags
30 Jun 2013, 17:47

Does anyone have any similar questions this this one?

I like this problem.

Thanks,

Hunter

_________________

4/28 GMATPrep 42Q 36V 640

Intern

Joined: 18 May 2013

Posts: 7

Location: Germany

GMAT Date : 09-27-2013

Followers: 0

Kudos [? ]:
9
[0 ] , given: 55

Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink ]

Show Tags
25 Jul 2013, 01:48

Bunuel wrote:

\((\frac{x+1}{x-1})^2\) If x#0 and x#1, and if x is replaced by 1/x everywhere in the expression above, then the resulting expression is equivalent to A. \((\frac{x+1}{x-1})^2\) B. \((\frac{x-1}{x+1})^2\) C. \(\frac{x^2+1}{1-x^2}\) D. \(\frac{x^2-1}{x^2+1}\) E. \(-(\frac{x-1}{x+1})^2\) \((\frac{\frac{1}{x}+1}{\frac{1}{x}-1})^2=(\frac{\frac{1+x}{x}}{\frac{1-x}{x}})^2=(\frac{1+x}{1-x})^2=(\frac{x+1}{x-1})^2\). Answer: A.

I don´t get \((\frac{\frac{1}{x}+1}{\frac{1}{x}-1})^2=(\frac{\frac{1+x}{x}}{\frac{1-x}{x}})^2\).. could you please explain your steps in a few words?

Math Expert

Joined: 02 Sep 2009

Posts: 37080

Followers: 7242

Kudos [? ]:
96313
[1 ]
, given: 10733

Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink ]

Show Tags
25 Jul 2013, 02:12
sv3n wrote:

Bunuel wrote:

\((\frac{x+1}{x-1})^2\) If x#0 and x#1, and if x is replaced by 1/x everywhere in the expression above, then the resulting expression is equivalent to A. \((\frac{x+1}{x-1})^2\) B. \((\frac{x-1}{x+1})^2\) C. \(\frac{x^2+1}{1-x^2}\) D. \(\frac{x^2-1}{x^2+1}\) E. \(-(\frac{x-1}{x+1})^2\) \((\frac{\frac{1}{x}+1}{\frac{1}{x}-1})^2=(\frac{\frac{1+x}{x}}{\frac{1-x}{x}})^2=(\frac{1+x}{1-x})^2=(\frac{x+1}{x-1})^2\). Answer: A.

I don´t get \((\frac{\frac{1}{x}+1}{\frac{1}{x}-1})^2=(\frac{\frac{1+x}{x}}{\frac{1-x}{x}})^2\).. could you please explain your steps in a few words?

Step by step:

\((\frac{\frac{1}{x}+1}{\frac{1}{x}-1})^2=(\frac{\frac{1+x}{x}}{\frac{1-x}{x}})^2\).

\((\frac{\frac{1+x}{x}}{\frac{1-x}{x}})^2=(\frac{1+x}{x}*\frac{x}{1-x})^2\)

\((\frac{1+x}{x}*\frac{x}{1-x})^2=(\frac{1+x}{1-x})^2\)

\((\frac{1+x}{1-x})^2=(\frac{x+1}{x-1})^2\)

Can you please tell me which step didn't you understand?

_________________

New to the Math Forum? Please read this: All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders ; 8. Overlapping Sets | PDF of Math Book ; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years. Collection of Questions: PS: 1. Tough and Tricky questions ; 2. Hard questions ; 3. Hard questions part 2 ; 4. Standard deviation ; 5. Tough Problem Solving Questions With Solutions ; 6. Probability and Combinations Questions With Solutions ; 7 Tough and tricky exponents and roots questions ; 8 12 Easy Pieces (or not?) ; 9 Bakers' Dozen ; 10 Algebra set. ,11 Mixed Questions , 12 Fresh Meat DS: 1. DS tough questions ; 2. DS tough questions part 2 ; 3. DS tough questions part 3 ; 4. DS Standard deviation ; 5. Inequalities ; 6. 700+ GMAT Data Sufficiency Questions With Explanations ; 7 Tough and tricky exponents and roots questions ; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!! ; 10 Number Properties set. , 11 New DS set. What are GMAT Club Tests ? Extra-hard Quant Tests with Brilliant Analytics

Intern

Joined: 18 May 2013

Posts: 7

Location: Germany

GMAT Date : 09-27-2013

Followers: 0

Kudos [? ]:
9
[0 ] , given: 55

Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink ]

Show Tags
25 Jul 2013, 04:10

I do not understand the first of these four steps. -.- I understand that 1/x+1 is the same as 1+x/x, but why have you done it? I only get it if I see the result and go backwards..

Math Expert

Joined: 02 Sep 2009

Posts: 37080

Followers: 7242

Kudos [? ]:
96313
[1 ]
, given: 10733

Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink ]

Show Tags
25 Jul 2013, 05:59

Intern

Joined: 18 May 2013

Posts: 7

Location: Germany

GMAT Date : 09-27-2013

Followers: 0

Kudos [? ]:
9
[0 ] , given: 55

Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink ]

Show Tags
25 Jul 2013, 06:10

Tried it several times again. I think I got it know.. thanks.

Intern

Joined: 27 Dec 2013

Posts: 35

Concentration: Finance, General Management

Followers: 0

Kudos [? ]:
11
[2 ]
, given: 29

Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink ]

Show Tags
10 May 2014, 05:50
2

This post received KUDOS

1

This post was BOOKMARKED

Walkabout wrote:

\((\frac{x+1}{x-1})^2\) If x#0 and x#1, and if x is replaced by 1/x everywhere in the expression above, then the resulting expression is equivalent to A. \((\frac{x+1}{x-1})^2\) B. \((\frac{x-1}{x+1})^2\) C. \(\frac{x^2+1}{1-x^2}\) D. \(\frac{x^2-1}{x^2+1}\) E. \(-(\frac{x-1}{x+1})^2\)

We can use a substitution method and process of elimination method to avoid the cumbersome calculations.

Say x = 2

x is replaced by 1/x. So, [(1/x + 1)/(1/x -1)] ^2 = [1+x/1-x] ^2

By substituting 2 we get, [(1+2)/(1-2)] ^2 = 9

Now, substitute x = 2 in the answer choices and an answer choice that gives the final answer as 9 is the correct answer.

Option A gives us 9.

Hence, A is the correct answer.

_________________

Kindly consider for kudos if my post was helpful!

Intern

Joined: 21 Apr 2014

Posts: 7

Followers: 0

Kudos [? ]:
10
[1 ]
, given: 1

Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink ]

Show Tags
10 May 2014, 06:19
1

This post received KUDOS

nikhil007 wrote:

How did 1-X in the denominator become X-1 in last step?

Rewrite the equation so that \(x\) appears as the first term in the equation:

\((1-x)^2 = (-x+1)^2\)

let's now rewrite the equation so that \(x\) is positive:

\((1-x)^2 = (-x+1)^2 = [ (-1) \cdot (x-1)]^2\)

the laws of exponents establish that \((a \cdot b)^n = a^n \cdot b^n\) which means that:

\((1-x)^2 = (-x+1)^2 = [ (-1) \cdot (x-1)]^2 = (-1)^2 \cdot (x-1)^2\)

notice that \((-1)^2 = -1 \cdot -1 = 1\) therefore:

\((1-x)^2 = (-x+1)^2 = [ (-1) \cdot (x-1)]^2 = (-1)^2 \cdot (x-1)^2 = 1 \cdot (x-1)^2 = (x-1)^2\)

Current Student

Joined: 08 Feb 2014

Posts: 206

Location: United States

Concentration: Finance

GMAT 1 : 650 Q39 V41

WE: Analyst (Commercial Banking)

Followers: 2

Kudos [? ]:
77
[0 ] , given: 145

Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink ]

Show Tags
20 Aug 2014, 13:50

Can't we just multiply the numerator and denominator by x, after substituing in (1/x)?

SVP

Status: The Best Or Nothing

Joined: 27 Dec 2012

Posts: 1858

Location: India

Concentration: General Management, Technology

WE: Information Technology (Computer Software)

Followers: 49

Kudos [? ]:
1986
[1 ]
, given: 193

Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink ]

Show Tags
21 Aug 2014, 00:20
1

This post received KUDOS

JackSparr0w wrote:

Can't we just multiply the numerator and denominator by x, after substituing in (1/x)?

We require to compute\(\frac{1}{x} + 1\) & \(\frac{1}{x} - 1\) before that

Refer Bunuel's method; done the very best

_________________

Kindly press "+1 Kudos" to appreciate

Intern

Joined: 29 Oct 2014

Posts: 25

Followers: 0

Kudos [? ]:
9
[0 ] , given: 14

If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink ]

Show Tags
09 Dec 2014, 03:15

\((\frac{\frac{1}{x}+1}{\frac{1}{x}-1})^2\). I just expanded the formula out like this: (1/x+y)(1/x+y)/(1/x+y)(1/x-y) =(x+y)/(x-y) = which is the same as answer choice A when squared (?) Understand the other way mentioned above too but want to check if this is an alternative or if it's incorrect.

Senior Manager

Joined: 28 Feb 2014

Posts: 295

Location: United States

Concentration: Strategy, General Management

Followers: 1

Kudos [? ]:
123
[1 ]
, given: 133

Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink ]

Show Tags
08 Jan 2015, 19:53
1

This post received KUDOS

You can also plug in numbers, such as x=2. Since x is being replaced by (1/x) we will replace x with (1/2). The original equation gives us a solution of 9. plugging in (1/2) into all of the answer solutions present us with A as the only correct answer.

Senior Manager

Status: Math is psycho-logical

Joined: 07 Apr 2014

Posts: 443

Location: Netherlands

GMAT Date : 02-11-2015

WE: Psychology and Counseling (Other)

Followers: 2

Kudos [? ]:
112
[0 ] , given: 169

Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink ]

Show Tags
20 Jan 2015, 08:03

This is what I did, but I am not sure if it is correct: [x+1]^2 / [x-1]^2 [(1/x)+1]^2 / [(1/x)-1]^2 [(1+x)/x]^2 / [(1-x)/x]^2 [(1+x)x]^2 / [(1-x)x]^2 (1+x)^2 / (1-x)^2 ANS A It would be easier to read alligned vertically. How do we do that?

GMAT Club Legend

Joined: 09 Sep 2013

Posts: 13914

Followers: 589

Kudos [? ]:
167
[0 ] , given: 0

Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink ]

Show Tags
09 Feb 2016, 09:12

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.

_________________

GMAT Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources

Manager

Joined: 18 Aug 2013

Posts: 154

Location: India

Concentration: Operations, Entrepreneurship

GMAT 1 : 640 Q48 V28

GPA: 3.92

WE: Operations (Transportation)

Followers: 3

Kudos [? ]:
43
[0 ] , given: 127

Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink ]

Show Tags
13 Feb 2016, 08:26

I replaced x with a value say 2 and then solved the whole problem.Got the answer correct. and its easy too without any confusion.

Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th
[#permalink ]
13 Feb 2016, 08:26

Go to page
1 2
Next
[ 23 posts ]

Similar topics
Author
Replies
Last post
Similar Topics:

6
x is replaced by 1 - x everywhere in the expression 1/x - 1/(1 - x)
Bunuel
7
17 Nov 2014, 11:44
24
In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk
Bunuel
10
05 Mar 2014, 01:20
37
In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk
mydreammba
21
26 Jan 2012, 03:15
10
In the Sequence x0, x1, x2, ..., xn, each term from x1 to xk
tonebeeze
10
14 Dec 2010, 16:40
1
x-1/x=?
apoorvasrivastva
5
20 Dec 2009, 09:08