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# If x#0 and x#1, and if x is replaced by 1/x everywhere in th

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If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink]

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27 Dec 2012, 06:45
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$$(\frac{x+1}{x-1})^2$$

If x#0 and x#1, and if x is replaced by 1/x everywhere in the expression above, then the resulting expression is equivalent to

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

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

C. $$\frac{x^2+1}{1-x^2}$$

D. $$\frac{x^2-1}{x^2+1}$$

E. $$-(\frac{x-1}{x+1})^2$$
[Reveal] Spoiler: OA
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Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink]

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27 Dec 2012, 06:56
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$$(\frac{x+1}{x-1})^2$$

If x#0 and x#1, and if x is replaced by 1/x everywhere in the expression above, then the resulting expression is equivalent to

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

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

C. $$\frac{x^2+1}{1-x^2}$$

D. $$\frac{x^2-1}{x^2+1}$$

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

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

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Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink]

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02 May 2013, 15:38
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Bunuel wrote:
[b]$$(\frac{x+1}{x-1})^2$$

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

How did 1-X in the denominator become X-1 in last step?
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Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink]

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02 May 2013, 21:57
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nikhil007 wrote:
Bunuel wrote:
[b]$$(\frac{x+1}{x-1})^2$$

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

How did 1-X in the denominator become X-1 in last step?

It's not (1-x) that became (x-1).
$$(1-x)^2$$ is simply rewritten as $$(x-1)^2$$.
Both $$(1-x)^2$$ and $$(x-1)^2$$ are essentially the same
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Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink]

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02 May 2013, 22:03
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nikhil007 wrote:
Bunuel wrote:
[b]$$(\frac{x+1}{x-1})^2$$

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

How did 1-X in the denominator become X-1 in last step?

$$(1-x)^2 = (x-1)^2$$. For example, $$(1-4)^2 = (4-1)^2 = 9$$.
The negative sign inside the bracket gets taken care of because of the square.
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Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink]

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30 Jun 2013, 18:47
Does anyone have any similar questions this this one?

I like this problem.

Thanks,
Hunter
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Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink]

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25 Jul 2013, 02:48
Bunuel wrote:
$$(\frac{x+1}{x-1})^2$$

If x#0 and x#1, and if x is replaced by 1/x everywhere in the expression above, then the resulting expression is equivalent to

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

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

C. $$\frac{x^2+1}{1-x^2}$$

D. $$\frac{x^2-1}{x^2+1}$$

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

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

I don´t get $$(\frac{\frac{1}{x}+1}{\frac{1}{x}-1})^2=(\frac{\frac{1+x}{x}}{\frac{1-x}{x}})^2$$.. could you please explain your steps in a few words?
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Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink]

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25 Jul 2013, 03:12
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Expert's post
sv3n wrote:
Bunuel wrote:
$$(\frac{x+1}{x-1})^2$$

If x#0 and x#1, and if x is replaced by 1/x everywhere in the expression above, then the resulting expression is equivalent to

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

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

C. $$\frac{x^2+1}{1-x^2}$$

D. $$\frac{x^2-1}{x^2+1}$$

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

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

I don´t get $$(\frac{\frac{1}{x}+1}{\frac{1}{x}-1})^2=(\frac{\frac{1+x}{x}}{\frac{1-x}{x}})^2$$.. could you please explain your steps in a few words?

Step by step:

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

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

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

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

Can you please tell me which step didn't you understand?
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Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink]

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25 Jul 2013, 05:10
I do not understand the first of these four steps. -.-

I understand that 1/x+1 is the same as 1+x/x, but why have you done it?
I only get it if I see the result and go backwards..
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Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink]

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25 Jul 2013, 06:59
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sv3n wrote:
I do not understand the first of these four steps. -.-

I understand that 1/x+1 is the same as 1+x/x, but why have you done it?
I only get it if I see the result and go backwards..

We want to simplify $$(\frac{\frac{1}{x}+1}{\frac{1}{x}-1})^2$$ so this step seemed quite natural to me.
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Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink]

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25 Jul 2013, 07:10
Tried it several times again. I think I got it know.. thanks.
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Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink]

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10 May 2014, 06:50
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$$(\frac{x+1}{x-1})^2$$

If x#0 and x#1, and if x is replaced by 1/x everywhere in the expression above, then the resulting expression is equivalent to

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

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

C. $$\frac{x^2+1}{1-x^2}$$

D. $$\frac{x^2-1}{x^2+1}$$

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

We can use a substitution method and process of elimination method to avoid the cumbersome calculations.

Say x = 2

x is replaced by 1/x. So, [(1/x + 1)/(1/x -1)] ^2 = [1+x/1-x] ^2

By substituting 2 we get, [(1+2)/(1-2)] ^2 = 9

Now, substitute x = 2 in the answer choices and an answer choice that gives the final answer as 9 is the correct answer.

Option A gives us 9.

Hence, A is the correct answer.
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Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink]

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10 May 2014, 07:19
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nikhil007 wrote:
How did 1-X in the denominator become X-1 in last step?

Rewrite the equation so that $$x$$ appears as the first term in the equation:

$$(1-x)^2 = (-x+1)^2$$

let's now rewrite the equation so that $$x$$ is positive:

$$(1-x)^2 = (-x+1)^2 = [ (-1) \cdot (x-1)]^2$$

the laws of exponents establish that $$(a \cdot b)^n = a^n \cdot b^n$$ which means that:

$$(1-x)^2 = (-x+1)^2 = [ (-1) \cdot (x-1)]^2 = (-1)^2 \cdot (x-1)^2$$

notice that $$(-1)^2 = -1 \cdot -1 = 1$$ therefore:

$$(1-x)^2 = (-x+1)^2 = [ (-1) \cdot (x-1)]^2 = (-1)^2 \cdot (x-1)^2 = 1 \cdot (x-1)^2 = (x-1)^2$$
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Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink]

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20 Aug 2014, 14:50
Can't we just multiply the numerator and denominator by x, after substituing in (1/x)?
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Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink]

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21 Aug 2014, 01:20
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JackSparr0w wrote:
Can't we just multiply the numerator and denominator by x, after substituing in (1/x)?

We require to compute$$\frac{1}{x} + 1$$ & $$\frac{1}{x} - 1$$ before that

Refer Bunuel's method; done the very best
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If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink]

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09 Dec 2014, 04:15
$$(\frac{\frac{1}{x}+1}{\frac{1}{x}-1})^2$$.

I just expanded the formula out like this:

(1/x+y)(1/x+y)/(1/x+y)(1/x-y)
=(x+y)/(x-y) = which is the same as answer choice A when squared (?)

Understand the other way mentioned above too but want to check if this is an alternative or if it's incorrect.
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Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink]

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08 Jan 2015, 20:53
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You can also plug in numbers, such as x=2. Since x is being replaced by (1/x) we will replace x with (1/2).

The original equation gives us a solution of 9.

plugging in (1/2) into all of the answer solutions present us with A as the only correct answer.
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Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink]

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20 Jan 2015, 09:03
This is what I did, but I am not sure if it is correct:

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

It would be easier to read alligned vertically. How do we do that?
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Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th [#permalink]

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13 Feb 2016, 09:26
I replaced x with a value say 2 and then solved the whole problem.Got the answer correct.
and its easy too without any confusion.
Re: If x#0 and x#1, and if x is replaced by 1/x everywhere in th   [#permalink] 13 Feb 2016, 09:26

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