Author
Message
TAGS:

Hide Tags
Manager

Joined: 02 Dec 2012

Posts: 178

Kudos [? ]:
3569 [0 ], given: 0

+1 Kudos

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

Show Tags
27 Dec 2012, 06:45

37

This post was BOOKMARKED

Question Stats:

70% (01:22) correct

30% (01:57) wrong

based on 1324 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 .

Kudos [? ]:
3569 [0 ], given: 0

+1 Kudos

Math Expert

Joined: 02 Sep 2009

Posts: 42248

Kudos [? ]:
132507 [7 ], given: 12323

+1 Kudos

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

Show Tags
27 Dec 2012, 06:56
7

This post received KUDOS

Expert's post

12

This post was BOOKMARKED

Kudos [? ]:
132507 [7 ], given: 12323

+1 Kudos

Manager

Joined: 04 Dec 2011

Posts: 80

Kudos [? ]:
29 [7 ], given: 13

+1 Kudos

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

Show Tags
02 May 2013, 15:38
7

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

Kudos [? ]:
29 [7 ], given: 13

+1 Kudos

Intern

Joined: 23 Apr 2013

Posts: 22

Kudos [? ]:
23 [8 ], given: 1

+1 Kudos

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

Show Tags
02 May 2013, 21:57
8

This post received KUDOS

4

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

Kudos [? ]:
23 [8 ], given: 1

+1 Kudos

Verbal Forum Moderator

Joined: 10 Oct 2012

Posts: 627

Kudos [? ]:
1384 [5 ], given: 136

+1 Kudos

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

Show Tags
02 May 2013, 22:03
5

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?

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

Kudos [? ]:
1384 [5 ], given: 136

+1 Kudos

Manager

Joined: 29 Mar 2010

Posts: 135

Kudos [? ]:
158 [0 ], given: 16

+1 Kudos

Location: United States

Concentration: Finance, International Business

GPA: 2.54

WE: Accounting (Hospitality and Tourism)

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

Show Tags
30 Jun 2013, 18:47

Does anyone have any similar questions this this one?

I like this problem.

Thanks,

Hunter

_________________

4/28 GMATPrep 42Q 36V 640

Kudos [? ]:
158 [0 ], given: 16

+1 Kudos

Intern

Joined: 18 May 2013

Posts: 7

Kudos [? ]:
10 [0 ], given: 55

+1 Kudos

Location: Germany

GMAT Date : 09-27-2013

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

Show Tags
25 Jul 2013, 02:48

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?

Kudos [? ]:
10 [0 ], given: 55

+1 Kudos

Math Expert

Joined: 02 Sep 2009

Posts: 42248

Kudos [? ]:
132507 [3 ], given: 12323

+1 Kudos

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

Show Tags
25 Jul 2013, 03:12
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: Ultimate GMAT Quantitative Megathread | 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

Kudos [? ]:
132507 [3 ], given: 12323

+1 Kudos

Intern

Joined: 18 May 2013

Posts: 7

Kudos [? ]:
10 [0 ], given: 55

+1 Kudos

Location: Germany

GMAT Date : 09-27-2013

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

Show Tags
25 Jul 2013, 05:10

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..

Kudos [? ]:
10 [0 ], given: 55

+1 Kudos

Math Expert

Joined: 02 Sep 2009

Posts: 42248

Kudos [? ]:
132507 [1 ], given: 12323

+1 Kudos

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

Show Tags
25 Jul 2013, 06:59
Kudos [? ]:
132507 [1 ], given: 12323

+1 Kudos

Intern

Joined: 18 May 2013

Posts: 7

Kudos [? ]:
10 [0 ], given: 55

+1 Kudos

Location: Germany

GMAT Date : 09-27-2013

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

Show Tags
25 Jul 2013, 07:10

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

Kudos [? ]:
10 [0 ], given: 55

+1 Kudos

Intern

Joined: 27 Dec 2013

Posts: 33

Kudos [? ]:
13 [2 ], given: 29

+1 Kudos

Concentration: Finance, General Management

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

Show Tags
10 May 2014, 06:50
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!

Kudos [? ]:
13 [2 ], given: 29

+1 Kudos

Intern

Joined: 21 Apr 2014

Posts: 7

Kudos [? ]:
16 [3 ], given: 1

+1 Kudos

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

Show Tags
10 May 2014, 07:19
3

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:

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

Kudos [? ]:
16 [3 ], given: 1

+1 Kudos

Current Student

Joined: 08 Feb 2014

Posts: 206

Kudos [? ]:
83 [0 ], given: 145

+1 Kudos

Location: United States

Concentration: Finance

WE: Analyst (Commercial Banking)

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

Show Tags
20 Aug 2014, 14:50

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

Kudos [? ]:
83 [0 ], given: 145

+1 Kudos

SVP

Status: The Best Or Nothing

Joined: 27 Dec 2012

Posts: 1852

Kudos [? ]:
2707 [1 ], given: 193

+1 Kudos

Location: India

Concentration: General Management, Technology

WE: Information Technology (Computer Software)

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

Show Tags
21 Aug 2014, 01:20
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

Kudos [? ]:
2707 [1 ], given: 193

+1 Kudos

Intern

Joined: 29 Oct 2014

Posts: 23

Kudos [? ]:
12 [0 ], given: 14

+1 Kudos

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

Show Tags
09 Dec 2014, 04:15

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

Kudos [? ]:
12 [0 ], given: 14

+1 Kudos

Senior Manager

Joined: 28 Feb 2014

Posts: 295

Kudos [? ]:
141 [1 ], given: 133

+1 Kudos

Location: United States

Concentration: Strategy, General Management

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

Show Tags
08 Jan 2015, 20:53
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.

Kudos [? ]:
141 [1 ], given: 133

+1 Kudos

Senior Manager

Status: Math is psycho-logical

Joined: 07 Apr 2014

Posts: 437

Kudos [? ]:
140 [0 ], given: 169

+1 Kudos

Location: Netherlands

GMAT Date : 02-11-2015

WE: Psychology and Counseling (Other)

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

Show Tags
20 Jan 2015, 09:03

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?

Kudos [? ]:
140 [0 ], given: 169

+1 Kudos

Manager

Joined: 18 Aug 2013

Posts: 150

Kudos [? ]:
57 [0 ], given: 127

+1 Kudos

Location: India

Concentration: Operations, Entrepreneurship

GPA: 3.92

WE: Operations (Transportation)

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

Show Tags
13 Feb 2016, 09:26

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.

Kudos [? ]:
57 [0 ], given: 127

+1 Kudos

Intern

Joined: 12 Apr 2015

Posts: 6

Kudos [? ]:
[0 ], given: 88

+1 Kudos

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

Show Tags
18 Aug 2016, 17:50

I was tripping over algebra until I realized that no math is needed at all here. You can take your time, triple-check, and solve this in < 30 seconds. It's simple algebraic substitution. Here's the key: This formula is valid for all X except 1 and 0. Notice that (A), the correct answer, is exactly the same formula as the question stem. We could just as easily substitute y, 1/W, pi, z^2, 3^(1/2) or 129,000 in for X and the formula will behave in exactly the same way. All you have to do is double-check that you're selecting the equivalent (in this case, identical) formula from the answer choices. To make this problem more difficult, they could have chosen an answer that is nothing more than an algebraically manipulated version of the original formula. So then, after carefully working through the algebra, you'd wind up with the same thing you started with and still have to massage it a bit in order to compare with the answer choices. I'll be looking out for that trick on G-Day. Someone please correct me if I'm wrong, but that appears to be the shortcut for this problem.

Kudos [? ]:
[0 ], given: 88

+1 Kudos

Re: If x ≠ 0 and x ≠ 1, and if x is replaced by 1/x everywhere in the
[#permalink ]
18 Aug 2016, 17:50

Go to page
1 2
Next
[ 27 posts ]