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If x ≠ 0 and x ≠ 1, and if x is replaced by 1/x everywhere in the

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

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New post 11 Jan 2017, 19:40
Is this problem testing the fact that multiplying or dividing both the numerator and the denominator of a fraction by the same number does not change the value of the fraction? In that case, a quick glance would show that the answer is A.
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Re: If x ≠ 0 and x ≠ 1, and if x is replaced by 1/x everywhere in the  [#permalink]

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New post 15 Feb 2017, 03:39
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For me the key to solving this question was to understand that \((1-x)^2=(x-1)^2\). To check that this is correct I tested a few cases: \((5-3)^2=4, (3-5)^2=4; (1-6)^2=25, (6-1)^2=25\).

Full solution: \((\frac{1}{x+1/1/x-1})^2=(\frac{1+x}{x/1-x/x})^2=(\frac{1+x}{1-x})^2=(\frac{x+1}{x-1})^2\)
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Re: If x ≠ 0 and x ≠ 1, and if x is replaced by 1/x everywhere in the  [#permalink]

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New post 09 Apr 2017, 21:38
Bunuel wrote:
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?


Hi Bunuel,

Regarding the last step: \({ \left( \cfrac { 1+x }{ 1-x } \right) }^{ 2 }={ \left( \cfrac { x+1 }{ x-1 } \right) }^{ 2 }\)

The reason why we can move the x and 1 in the numerator is because they're positive, so it's indifferent which one comes first, correct?
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Re: If x ≠ 0 and x ≠ 1, and if x is replaced by 1/x everywhere in the  [#permalink]

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New post 09 Apr 2017, 22:53
diegocml wrote:
Bunuel wrote:
sv3n wrote:
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?


Hi Bunuel,

Regarding the last step: \({ \left( \cfrac { 1+x }{ 1-x } \right) }^{ 2 }={ \left( \cfrac { x+1 }{ x-1 } \right) }^{ 2 }\)

The reason why we can move the x and 1 in the numerator is because they're positive, so it's indifferent which one comes first, correct?


No.

For the numerator: changing the order of the addends does not change the sum: a + b = b + a. So, 1 + x = x + 1.

For the denominator: (a - b)^2 = (b - a)^2, so (1 - x)^2 = (x - 1)^2.
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Re: If x ≠ 0 and x ≠ 1, and if x is replaced by 1/x everywhere in the  [#permalink]

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New post 24 Aug 2017, 06:33
Hi,
I have a query if someone can explain.
I get ans as (1+x/1-x)^2. Options which confuse me are
B) and E)

Ans 1+x/1-x if both numerator and denominator is multiplied by - then we get (x-1/x+1) (B) and negative cancel out
or -(x-1/x+1)(E) negative stays... I am not sure. Can we multiply by a -ve sign when there is a bracket with a square.
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Re: If x ≠ 0 and x ≠ 1, and if x is replaced by 1/x everywhere in the  [#permalink]

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New post 06 Sep 2017, 06:14
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. \(-(\frac{x-1}{x+1})^2\)



Hey there - I have been struggling understanding questions in this format. Is there a guide anywhere for the same? Or is there a better browser to be used to read questions with fractions and powers?
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Re: If x ≠ 0 and x ≠ 1, and if x is replaced by 1/x everywhere in the  [#permalink]

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New post 06 Sep 2017, 06:19
zoompastthisGMAT wrote:
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. \(-(\frac{x-1}{x+1})^2\)



Hey there - I have been struggling understanding questions in this format. Is there a guide anywhere for the same? Or is there a better browser to be used to read questions with fractions and powers?


Check Algebra chapter in Ultimate GMAT Quantitative Megathread.

Hope it helps.
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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.


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

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New post 12 Oct 2018, 22:24
VeritasKarishma could you please share your approach here specially how it becomes from (1+x/1-x)^2 to (x+1/x-1)^2 ? Thanks.
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Re: If x ≠ 0 and x ≠ 1, and if x is replaced by 1/x everywhere in the   [#permalink] 12 Oct 2018, 22:24

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