If n 2 and (2/n) is substituted for all instances of n in

Question

(n – 2)^{-1}* (2 + n)

If n >2 and 2/n is substituted for all instances of n in the above expression, then the new expression will be equivalent to which of the following:

A.  (n + 1)(n – 1)^{(-1)} 
B.  –(n + 1)(n – 1)^{(-1)} 
C.  –(n – 1)(n + 1)^{(-1)} 
D.  (2 + n)^{(-1)}*(n – 2) 
E.  (n – 2)^{(-1)}*(2 + n)

Bunuel · Answer

Number plugging:

Say  n=3 , then substitute  \frac{2}{n}=\frac{2}{3}  into  (n - 2)^{-1} (2 + n)=-2 .

Now, substitute  n=3  into the answer choices and see which of them gives -2. Only B fits.

Answer: B.

Note that for plug-in method it might happen that for some particular number(s) more than one option may give "correct" answer. In this case just pick some other numbers and check again these "correct" options only.

Alebra:

Substitute  \frac{2}{n}  into  (n - 2)^{-1} (2 + n) :

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

Answer: B.

evdo · Answer

\frac{(n+1)}{(1-n)}=-\frac{(n+1)}{(n-1)}

How do you do this kind of operation? Can you share any study materials on this topic? Thanks

KarishmaB · Answer

This is basic algebra and fraction.

You take our "-" common from the denominator to get

\frac{(n+1)}{(1-n)}=\frac{(n+1)}{-(n-1)}

And from theory of fractions, you know that

\frac{(n+1)}{-(n-1)}=-\frac{(n+1)}{(n-1)} = \frac{-(n+1)}{(n-1)}

For these basics, check purplemath.com or khanacademy.org

mbaapp1234 · Answer

Picking n = 3 is the way to go for this problem. If n=3, then the formula simplifies to -2 when 2/n replaces n.

Then it's about neatness and working carefully. Start with answer choice A: the target value = 2, not -2. Eliminate A.

Move on to B. -(3+1)(3-1)^-1 = -4/(2)^-1 = -4/2 = -2. At this point, you should test the rest of the choices to ensure that there are no other answer choices that work for n = 3.

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If n >2 and (2/n) is substituted for all instances of n in

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