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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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05 Aug 2018, 00:34
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75% (hard)

Question Stats:

60% (02:17) correct 40% (01:57) wrong based on 62 sessions

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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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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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05 Aug 2018, 01:05
1
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.

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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05 Aug 2018, 01:32
1
delta23 wrote:

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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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!
Senior Manager
Joined: 22 Feb 2018
Posts: 257
Re: If x is an integer and x2(x−1)+2x+1=2x1+x+12(1−x)  [#permalink]

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05 Aug 2018, 05:16
1
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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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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Joined: 22 Feb 2018
Posts: 257
Re: If x is an integer and x2(x−1)+2x+1=2x1+x+12(1−x)  [#permalink]

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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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Posts: 3
Re: If x is an integer and x2(x−1)+2x+1=2x1+x+12(1−x)  [#permalink]

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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!
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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