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If 1/(x - 2) = 1/(x + 2) + 1/(x - 1), which of the following is a

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If 1/(x - 2) = 1/(x + 2) + 1/(x - 1), which of the following is a  [#permalink]

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New post 17 Nov 2014, 11:37
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A
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Tough and Tricky questions: Algebra.



If \(\frac{1}{x - 2} = \frac{1}{x + 2} + \frac{1}{x - 1}\), which of the following is a possible value of \(x\)?

A. -2
B. -1
C. 0
D. 1
E. 2

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Re: If 1/(x - 2) = 1/(x + 2) + 1/(x - 1), which of the following is a  [#permalink]

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New post 17 Nov 2014, 20:39
3
Answer = C = 0

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

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

\(\frac{1}{x-1} = \frac{x+2-(x-2)}{x^2 - 4}\)

\(\frac{1}{x-1} = \frac{4}{x^2 - 4}\)

\(x^2 - 4 = 4x - 4\)

x = 0 OR x = 2

If we take x = 2; LHS becomes void

hence x = 0
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Re: If 1/(x - 2) = 1/(x + 2) + 1/(x - 1), which of the following is a  [#permalink]

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New post 17 Nov 2014, 23:03
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Bunuel wrote:

Tough and Tricky questions: Algebra.



If \(\frac{1}{x - 2} = \frac{1}{x + 2} + \frac{1}{x - 1}\), which of the following is a possible value of \(x\)?

A. -2
B. -1
C. 0
D. 1
E. 2

Kudos for a correct solution.


Answer = C.
IMO simple way to look at this is if x = 2, -2 or 1, then one of the fractions listed becomes invalid. So none of these can be answers. That leaves us with -1 or 0.
Plug in each of these. I started with 0 first as it is easiest
LHS = -1/2
RHS = 1/2-1/1 = -1/2.
LHS = RHS, hence x=0.
Answer: C.
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Re: If 1/(x - 2) = 1/(x + 2) + 1/(x - 1), which of the following is a  [#permalink]

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New post 18 Nov 2014, 01:59
1
Just put the value of x from the given options and check
we can verify equation is valid only for x=0

Hence C
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Re: If 1/(x - 2) = 1/(x + 2) + 1/(x - 1), which of the following is a  [#permalink]

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New post 18 Nov 2014, 06:58
Official Solution:

If \(\frac{1}{x - 2} = \frac{1}{x + 2} + \frac{1}{x - 1}\), which of the following is a possible value of \(x\)?

A. -2
B. -1
C. 0
D. 1
E. 2

The fastest way to solve this problem is first to recognize that an algebraic approach will take a little time. Essentially, we will have to multiply through by the product \((x - 2)(x + 2)(x - 1)\), then simplify.

If, instead, we glance at the answer choices, we see that 3 of them make one of the denominators zero, a result that is not allowed (we cannot divide by zero). Specifically, \(x\) cannot be -2 because one denominator is \(x + 2\); likewise, \(x\) cannot be 1 or 2, since we have \(x - 1\) and \(x - 2\) as denominators as well.

Thus, the only two possible answers are -1 and 0. We try each in turn.

If \(x = -1\), then we have the following:

\(\frac{1}{-3} = \frac{1}{1} + \frac{1}{-2}\)?

\(-\frac{1}{3} = 1 - \frac{1}{2}\)?

This is not true.

However, if \(x = 0\), then we have the following:

\(\frac{1}{-2} = \frac{1}{2} + \frac{1}{-1}\)?

\(-\frac{1}{2} = \frac{1}{2} - 1\)?

\(-\frac{1}{2} = -\frac{1}{2}\)?

This is true, so \(x\) can be equal to 0.

Alternatively, we could take the algebraic approach.

First, we multiply through by the product \((x - 2)(x + 2)(x - 1)\) to eliminate denominators.
\((x - 1)(x + 2) = (x - 2)(x - 1) + (x - 2)(x + 2)\)
\(x^2 + x - 2 = x^2 - 3x + 2 + x^2 - 4\)
\(0 = x^2 - 4x\)
\(0 = x(x - 4)\)

\(x = 0\) or \(x = 4\)

Answer: C.
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Collection of Questions:
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Re: If 1/(x - 2) = 1/(x + 2) + 1/(x - 1), which of the following is a  [#permalink]

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New post 28 Mar 2017, 06:30
There is always 2 method to solve these question in most cases either putting values or solving equation...
I need expert opinion which is the fastest way to do it
I usually find putting values fastest not sure about precision what you people say?
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Re: If 1/(x - 2) = 1/(x + 2) + 1/(x - 1), which of the following is a  [#permalink]

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New post 28 Mar 2017, 11:48
An easy approach to solve the problem is process elimination.
If we use 0 for x

1/0 - 2 = 1/0+2 + 1/0-1
1/-2 = 1/2 + 1/-1
-1/2 = -1/2

Try the other answers and you will see the answers do not coincide.

The answer is C
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Re: If 1/(x - 2) = 1/(x + 2) + 1/(x - 1), which of the following is a  [#permalink]

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New post 31 Mar 2017, 08:10
Bunuel wrote:

Tough and Tricky questions: Algebra.



If \(\frac{1}{x - 2} = \frac{1}{x + 2} + \frac{1}{x - 1}\), which of the following is a possible value of \(x\)?

A. -2
B. -1
C. 0
D. 1
E. 2

Kudos for a correct solution.


By substituting the value we 0 will satisfy the given equation
Hence C
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Re: If 1/(x - 2) = 1/(x + 2) + 1/(x - 1), which of the following is a  [#permalink]

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Re: If 1/(x - 2) = 1/(x + 2) + 1/(x - 1), which of the following is a   [#permalink] 08 Oct 2018, 05:08
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