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

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

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

Answer: A.
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How did 1-X in the denominator become X-1 in last step?

Does anyone have any similar questions this this one?

I like this problem.

Thanks,
Hunter
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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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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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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.

Can't we just multiply the numerator and denominator by x, after substituing in (1/x)?

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

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.

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?

I was tripping over algebra until I realized that no math is needed at all here. You can take your time, triple-check, and solve this in < 30 seconds.

It's simple algebraic substitution. Here's the key: This formula is valid for all X except 1 and 0. Notice that (A), the correct answer, is exactly the same formula as the question stem.

We could just as easily substitute y, 1/W, pi, z^2, 3^(1/2) or 129,000 in for X and the formula will behave in exactly the same way. All you have to do is double-check that you're selecting the equivalent (in this case, identical) formula from the answer choices.

To make this problem more difficult, they could have chosen an answer that is nothing more than an algebraically manipulated version of the original formula. So then, after carefully working through the algebra, you'd wind up with the same thing you started with and still have to massage it a bit in order to compare with the answer choices. I'll be looking out for that trick on G-Day.

Someone please correct me if I'm wrong, but that appears to be the shortcut for this problem.