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If x is an integer and x2(x−1)+2x+1=2x1+x+12(1−x)

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If x is an integer and x2(x−1)+2x+1=2x1+x+12(1−x)  [#permalink]

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New post 05 Aug 2018, 00:34
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Question Stats:

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If x is an integer and \(\frac{x}{2(x−1)}\)+\(\frac{2}{x+1}\)=\(\frac{2x}{1+x}\)+\(\frac{1}{2(1−x)}\), what is the value of x?


A) -3

B) -1

C) 0

D) 2

E) 3

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If x is an integer and x2(x−1)+2x+1=2x1+x+12(1−x)  [#permalink]

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New post 05 Aug 2018, 00:57
delta23 wrote:
If x is an integer and \(\frac{x}{2(x−1)}\)+\(\frac{2}{x+1}\)=\(\frac{2x}{1+x}\)+\(\frac{1}{2(1−x)}\), what is the value of x?


A) -3

B) -1

C) 0

D) 2

E) 3


\(\frac{x}{2(x−1)}\)+\(\frac{2}{x+1}\)=\(\frac{2x}{1+x}\)+\(\frac{1}{2(1−x)}\)
Or, \(\frac{x(x+1)+4(x-1)}{2(x+1)(x-1)}=\frac{4x(1-x)+(1+x)}{2(1+x)(1-x)}\)
Or, \(\frac{x^2+5x-4}{2(x+1)(x-1)}= \frac{-4x^2+5x+1}{-2(x+1)(x-1)}\)
Or,\(x^2+5x-4=-(-4x^2+5x+1)\)
Or, \(x^2+5x-4=4x^2-5x-1\)
Or, \(3x^2-9x-x+3=0\)
Or, 3x(x-3)-1(x-3)=0
Or, (x-3)(3x-1)=0
Or, x=3 or, x=\(\frac{1}{3}\)

Given,x is an integer. Hence, x=3.

Ans. (E)
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Re: If x is an integer and x2(x−1)+2x+1=2x1+x+12(1−x)  [#permalink]

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New post 05 Aug 2018, 01:05
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delta23 wrote:
If x is an integer and \(\frac{x}{2(x−1)}\)+\(\frac{2}{x+1}\)=\(\frac{2x}{1+x}\)+\(\frac{1}{2(1−x)}\), what is the value of x?


A) -3

B) -1

C) 0

D) 2

E) 3


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

Plugging in options would be my first choice here. Not a lot of work is required to try x = 0 since some terms become 0.
x cannot be -1 since we can't have 0 in the denominator.
2 is possible option but that doesn't work.
Try x = 3 and it fits.

Answer (E)

Note that you can simplify the terms by bringing the common terms together and then plug in but as it is too it is not cumbersome.
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Re: If x is an integer and x2(x−1)+2x+1=2x1+x+12(1−x)  [#permalink]

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New post 05 Aug 2018, 01:32
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delta23 wrote:
KarishmaB Bunuel, please help


Easiest way to do it is to plug answer choices in and see which one works. Only E works.
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If x is an integer and x2(x−1)+2x+1=2x1+x+12(1−x)  [#permalink]

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New post 05 Aug 2018, 04:53
Would there be a way to solve this in under 2 minutes?

I tried simplifying the equation to reach the answer and it took me approx. 7 minutes to do so.

Thanks!
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Re: If x is an integer and x2(x−1)+2x+1=2x1+x+12(1−x)  [#permalink]

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New post 05 Aug 2018, 05:16
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WIKI18 wrote:
Would there be a way to solve this in under 2 minutes?

I tried simplifying the equation to reach the answer and it took me approx. 7 minutes to do so.

Thanks!


OA: E

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

Multiplying both sides by \(2(x-1)(x+1)\),we get

\(x(x+1)+4(x-1)=4x(x-1)-(x+1)\)

\(x^2+x+4x-4=4x^2-4x-x-1\)

\(3x^2 -10x+3=0\)
\(3x^2 -9x-1x+3=0\)
\(3x(x-3)-1(x-3)=0\)
\((3x-1)(x-3)=0\)
\(x=\frac{1}{3}\)(Rejected as \(x\) should be an integer), \(3\)
\(x=3\)
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Re: If x is an integer and x2(x−1)+2x+1=2x1+x+12(1−x)  [#permalink]

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New post 05 Aug 2018, 05:38
Princ wrote:
WIKI18 wrote:
Would there be a way to solve this in under 2 minutes?

I tried simplifying the equation to reach the answer and it took me approx. 7 minutes to do so.

Thanks!


OA: E

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

Multiplying both sides by \(2(x-1)(x+1)\),we get

\(x(x+1)+4(x-1)=4x(x-1)-(x+1)\)

\(x^2+x+4x-4=4x^2-4x-x-1\)

\(3x^2 -10x+3=0\)
\(3x^2 -9x-1x+3=0\)
\(3x(x-3)-1(x-3)=0\)
\((3x-1)(x-3)=0\)
\(x=\frac{1}{3}\)(Rejected as \(x\) should be an integer), \(3\)
\(x=3\)


When I multiply the last term i.e. \(\frac{1}{2(1-x)}\) by \(2(x-1)(x+1)\) I get \(\frac{(x-1)(x+1)}{(1-x)}\)

Can you please explain how you reached\((x+1)\)?
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Re: If x is an integer and x2(x−1)+2x+1=2x1+x+12(1−x)  [#permalink]

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New post 05 Aug 2018, 07:00
1
WIKI18 wrote:
Princ wrote:
WIKI18 wrote:
Would there be a way to solve this in under 2 minutes?

I tried simplifying the equation to reach the answer and it took me approx. 7 minutes to do so.

Thanks!


OA: E

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

Multiplying both sides by \(2(x-1)(x+1)\),we get

\(x(x+1)+4(x-1)=4x(x-1)-(x+1)\)

\(x^2+x+4x-4=4x^2-4x-x-1\)

\(3x^2 -10x+3=0\)
\(3x^2 -9x-1x+3=0\)
\(3x(x-3)-1(x-3)=0\)
\((3x-1)(x-3)=0\)
\(x=\frac{1}{3}\)(Rejected as \(x\) should be an integer), \(3\)
\(x=3\)


When I multiply the last term i.e. \(\frac{1}{2(1-x)}\) by \(2(x-1)(x+1)\) I get \(\frac{(x-1)(x+1)}{(1-x)}\)

Can you please explain how you reached\((x+1)\)?

WIKI18
\(2(x-1)(x+1)\)*\(\frac{1}{2(1-x)}\)=\(2(x-1)(x+1)\)*\(\frac{1}{-2(x-1)}\) =\((x+1)\)*\(-1=-(x+1)\)
Let me know if further explaination is required.
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Re: If x is an integer and x2(x−1)+2x+1=2x1+x+12(1−x)  [#permalink]

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New post 05 Aug 2018, 07:27
Princ wrote:

\(2(x-1)(x+1)\)*\(\frac{1}{2(1-x)}\)=\(2(x-1)(x+1)\)*\(\frac{1}{-2(x-1)}\) =\((x+1)\)*\(-1=-(x+1)\)
Let me know if further explaination is required.


Got it now. Thanks!
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Re: If x is an integer and x2(x−1)+2x+1=2x1+x+12(1−x) &nbs [#permalink] 05 Aug 2018, 07:27
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