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# If x^2 ≠ 1, which is equal to x/(x + 1) − x/(x − 1) + 2/(x^2 − 1) ?

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Joined: 02 Sep 2009
Posts: 52161
If x^2 ≠ 1, which is equal to x/(x + 1) − x/(x − 1) + 2/(x^2 − 1) ?  [#permalink]

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20 Apr 2017, 22:19
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45% (medium)

Question Stats:

74% (02:12) correct 26% (02:52) wrong based on 73 sessions

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If x^2 ≠ 1, which is equal to $$\frac{x}{(x + 1)} − \frac{x}{(x − 1)} + \frac{2}{(x^2 − 1)}$$?

A. $$\frac{(−x + 3)}{(x + 1)}$$

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

C. $$\frac{−2}{(x + 1)}$$

D. $$\frac{2}{(x − 1)}$$

E. $$\frac{−2}{(x − 1)}$$

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Joined: 24 Apr 2016
Posts: 331
Re: If x^2 ≠ 1, which is equal to x/(x + 1) − x/(x − 1) + 2/(x^2 − 1) ?  [#permalink]

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21 Apr 2017, 00:53
1
x/x+1 - x/x-1 + 2/(x^2 -1)

Above can be written as

==> [x(x-1)/(x^2 -1] - [x(x+1)/(x^2 -1] + 2/(x^2 -1)

==> [x^2 -x -x^2 -x +2](x^2 -1)

==> [2-2x]/(x+1)(x-1)

==> -2(x-1)/(x+1)(x-1)

==> -2/(x+1)

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Re: If x^2 ≠ 1, which is equal to x/(x + 1) − x/(x − 1) + 2/(x^2 − 1) ?  [#permalink]

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25 Apr 2017, 08:11
Bunuel wrote:
If x^2 ≠ 1, which is equal to $$\frac{x}{(x + 1)} − \frac{x}{(x − 1)} + \frac{2}{(x^2 − 1)}$$?

A. $$\frac{(−x + 3)}{(x + 1)}$$

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

C. $$\frac{−2}{(x + 1)}$$

D. $$\frac{2}{(x − 1)}$$

E. $$\frac{−2}{(x − 1)}$$

Let’s get each fraction in terms of the common denominator of (x + 1)(x - 1).

[x(x - 1)]/[(x - 1)(x + 1)] - [x(x + 1)]/[(x - 1)(x + 1)] + 2/(x^2 + 1)

(x^2 - x - x^2 - x + 2)/[(x + 1)(x - 1)]

(-2x + 2)/[(x + 1)(x - 1)]

-2(x - 1)/[(x + 1)(x - 1)]

-2/(x + 1)

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Math Expert
Joined: 02 Aug 2009
Posts: 7199
If x^2 ≠ 1, which is equal to x/(x + 1) − x/(x − 1) + 2/(x^2 − 1) ?  [#permalink]

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25 Apr 2017, 08:30
Bunuel wrote:
If x^2 ≠ 1, which is equal to $$\frac{x}{(x + 1)} − \frac{x}{(x − 1)} + \frac{2}{(x^2 − 1)}$$?

A. $$\frac{(−x + 3)}{(x + 1)}$$

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

C. $$\frac{−2}{(x + 1)}$$

D. $$\frac{2}{(x − 1)}$$

E. $$\frac{−2}{(x − 1)}$$

Hi..

Another method apart from the one choosen above.
May be good for those find algebra difficult..
substitute x as 2 & find value in equation and match with choices

$$\frac{x}{(x + 1)} − \frac{x}{(x − 1)} + \frac{2}{(x^2 − 1)}= \frac{2}{(2 + 1)} − \frac{2}{(2 − 1)} + \frac{2}{(2^2 − 1)}= \frac{2}{3} − 2 + \frac{2}{3}=\frac{-2}{3}$$

Let's check the choices..

A. $$\frac{(−x + 3)}{(x + 1)}=\frac{1}{3}$$... NO

B. $$\frac{2}{(x + 1)}=\frac{2}{3}$$.. NO

C. $$\frac{−2}{(x + 1)}=\frac{-2}{3}$$... YES

D. $$\frac{2}{(x − 1)}=2$$... NO

E. $$\frac{−2}{(x − 1)}=-2$$.. NO

C
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

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Joined: 02 Aug 2013
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Location: India
WE: Programming (Consulting)
Re: If x^2 ≠ 1, which is equal to x/(x + 1) − x/(x − 1) + 2/(x^2 − 1) ?  [#permalink]

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27 Apr 2017, 10:31
1
$$\frac{{x(x+1) - x(x-1) +2 }}{{(x^2 - 1)}}$$

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

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

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

$$\frac{{-2}}{{(x+1)}}$$

Ans: Option C
Re: If x^2 ≠ 1, which is equal to x/(x + 1) − x/(x − 1) + 2/(x^2 − 1) ? &nbs [#permalink] 27 Apr 2017, 10:31
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