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Given that ∣x∣≠1, the expression 1/(x+1)+1/(x−1) is equivalent to whic

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Bunuel
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Given that ∣x∣≠1, the expression 1/(x+1)+1/(x−1) is equivalent to whic [#permalink] New post  20 Feb 2017, 02:34
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Given that ∣x∣≠1, the expression \(\frac{1}{(x+1)}+\frac{1}{(x−1)}\) is equivalent to which of the following?

A. 1/x

B. (2x+1)/(x^2−1)

C. 2x

D. 2x/(x^2−1)

E. 2x^2−1
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Re: Given that ∣x∣≠1, the expression 1/(x+1)+1/(x−1) is equivalent to whic [#permalink] New post  20 Feb 2017, 02:52
=(x-1+x+1)/(x-1)(x+1)=2x/(x^2-1)
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Re: Given that ∣x∣≠1, the expression 1/(x+1)+1/(x−1) is equivalent to whic [#permalink] New post  20 Feb 2017, 03:21
1/(x+1) + 1/(x-1) = (x-1 + x-1)/(x+1)*(x-1) = 2x/(x^2-1);
Answer D.

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BrentGMATPrepNow
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Given that ∣x∣≠1, the expression 1/(x+1)+1/(x−1) is equivalent to whic [#permalink] New post  Updated on: 25 May 2020, 12:54
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Bunuel wrote:
Given that ∣x∣≠1, the expression \(\frac{1}{(x+1)}+\frac{1}{(x−1)}\) is equivalent to which of the following?

A. 1/x

B. (2x+1)/(x² - 1)

C. 2x

D. 2x/(x² - 1)

E. 2x² - 1


To combine (add) the two fractions, we need a common denominator.

So, take \(\frac{1}{(x+1)}\) and create an equivalent fraction by multiplying top and bottom by (x - 1) to get: \(\frac{(x - 1)}{(x² - 1)}\)

ASIDE: (x + 1)(x - 1) = x² - 1

Then take \(\frac{1}{(x-1)}\) and create an equivalent fraction by multiplying top and bottom by (x + 1) to get: \(\frac{(x + 1)}{(x² - 1)}\)

So, \(\frac{1}{(x+1)}+\frac{1}{(x−1)}\) = \(\frac{(x - 1)}{(x² - 1)}\) + \(\frac{(x + 1)}{(x² - 1)}\)

= \(\frac{2x}{(x² - 1)}\)

Answer: D

Originally posted by BrentGMATPrepNow on 20 Feb 2017, 09:51.
Last edited by BrentGMATPrepNow on 25 May 2020, 12:54, edited 1 time in total.
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Re: Given that ∣x∣≠1, the expression 1/(x+1)+1/(x−1) is equivalent to whic [#permalink] New post  21 Feb 2017, 03:17
Bunuel wrote:
Given that ∣x∣≠1, the expression \(\frac{1}{(x+1)}+\frac{1}{(x−1)}\) is equivalent to which of the following?

A. 1/x

B. (2x+1)/(x^2−1)

C. 2x

D. 2x/(x^2−1)

E. 2x^2−1



Another solution Put x =2, then the expression = 1/3 +1 =4/3

We easily can eliminate A, C & E

Apply in B, we get 5/3

we left with one answer.

Answer: D
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Re: Given that ∣x∣≠1, the expression 1/(x+1)+1/(x−1) is equivalent to whic [#permalink] New post  21 Feb 2017, 05:11
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Bunuel wrote:
Given that ∣x∣≠1, the expression \(\frac{1}{(x+1)}+\frac{1}{(x−1)}\) is equivalent to which of the following?

A. 1/x

B. (2x+1)/(x^2−1)

C. 2x

D. 2x/(x^2−1)

E. 2x^2−1


Since we have to take the LCM of the two fractions so denominator of the solution much have (x+1)*(x-1) which is (x^2-1) which is available only in Option B and D so Option A, C and E ruled out

Numerator should be the simple addition of (x+1) and (x-1) which is 2x hence

Answer: Option D

Although substituting the value of x ad checking options also is an easier way but it's always best to solve algeberically till you are stuck and want to substitute values to get out of mess
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Re: Given that ∣x∣≠1, the expression 1/(x+1)+1/(x−1) is equivalent to whic [#permalink] New post  21 Feb 2017, 05:17
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Bunuel wrote:
Given that ∣x∣≠1, the expression \(\frac{1}{(x+1)}+\frac{1}{(x−1)}\) is equivalent to which of the following?

A. 1/x

B. (2x+1)/(x^2−1)

C. 2x

D. 2x/(x^2−1)

E. 2x^2−1


Since we have to take the LCM of the two fractions so denominator of the solution much have (x+1)*(x-1) which is (x^2-1) which is available only in Option B and D so Option A, C and E ruled out

Numerator should be the simple addition of (x+1) and (x-1) which is 2x hence

Answer: Option D

Although substituting the value of x ad checking options also is an easier way but it's always best to solve algeberically till you are stuck and want to substitute values to get out of mess
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Re: Given that ∣x∣≠1, the expression 1/(x+1)+1/(x−1) is equivalent to whic [#permalink] New post  23 Feb 2017, 10:50
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Bunuel wrote:
Given that ∣x∣≠1, the expression \(\frac{1}{(x+1)}+\frac{1}{(x−1)}\) is equivalent to which of the following?

A. 1/x

B. (2x+1)/(x^2−1)

C. 2x

D. 2x/(x^2−1)

E. 2x^2−1


Let’s first get a common denominator of (x + 1)(x -1) by multiplying the first expression by (x -1)/(x -1) and the second expression by (x + 1)/(x + 1):

(x -1)/[(x -1)(x +1)] + (x + 1)/[(x -1)(x+1)]

[(x - 1) + (x + 1)]/[(x -1)(x + 1)]

2x/(x^2 - 1)

Answer: D
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Re: Given that ∣x∣≠1, the expression 1/(x+1)+1/(x−1) is equivalent to whic [#permalink]
23 Feb 2017, 10:50
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