Variables In Choices Problem

[jsnipes]

I am having difficulty with this question:

(n - 2)^-1(2 + n)

If n >2 and n/2 is substituted for all instances of n in the above expression, then the new expression will be equivalent to which of the following:
(n + 1)(n - 1)^-1
-(n + 1)(n - 1)^-1
-(n - 1)(n + 1)^-1
(2 + n)^-1(n - 2)
(n - 2)^-1(2 + n)

and the explanation that was provided says...

This problem is an algebra problem that features VICs, so we can attack it multiple ways. Although we could use Direct Algebra, the complexity of the substitution combined with the negative exponents provides a hint that using Direct Algebra might take more than two minutes. Instead, we can Pick Numbers and Calculate a Target because we can then plug our choice for n directly into the answer choices and quickly evaluate all of them in less than 2 minutes.

If anyone has any tips for how to quickly determine that this would be the case I am all ears because it doesn't seem to me that ordinarily plugging in an answer for 5 possible answers is going to be very quick.

Additionally, if someone could tell me why my algebraic answer is wrong and what the right one is that would be awesome.

So, first I plugged in (2/n) for every n as the problem stated. This gave me
[(2/n) -2]^-1*[2+(2/n)] or

[2+(2/n)]/[(2/n)-2]

[(2n+2)/n]/[(2-2n)/n]

(2n^2+2n)/(2n-2n^2)

[2n(n+1)]/[2n(1-n)]

(n+1)/(1-n) OR (n+1)(1-n)^-1

But that answer is nowhere to be found and wrong I guess. Help?

[GmatMathPro]

remember that (1-n) can be flipped around by factoring out a negative: (1-n)=-(n-1). That would make (1+n)/(1-n) equivalent to (n+1)/-(n-1). You're free to move the negative sign to the numerator, which would give you -(n+1)(n-1)^-1.

So it's not that you did anything wrong, you just have to be able to recognize the equivalent expression. However, your algebraic solution could be simpler. Once you get to the second step,(2+2/n)/(2/n-2), just multiply the top and bottom of the whole fraction by n to get rid of the fractions in the top and bottom. Then cancel out a 2, find the equivalent form, and you're done.

[jsnipes]

Thanks for the reply. That makes sense. So many little rules like that one that I have just completely forgot and keep popping up.

[DHP1985]

What I don't understand here, is that the question states substitute n/2 and you plugged in 2/n, please explain =) Did you not come to the conclusion of -1 because you have to expland? when I do this problem I get -1:
(n-2)^-1(2+n)
=(n/2-2)^-1(2+n/s)
=(2+n/s)/(n/2-2)-flip the negative exponent into the denominator
=n/s cancels out, thus 2/-2 which results in -1.

Clearly I have missed some factoring rule as -1 is not close to any of the options above.

[GmatMathPro]

Yeah I think that was just a typo he made when he posted the problem originally. It should be 2/n to get the correct answer.

However, be aware that you are making a critical mistake in your algebra. You cannot cancel out parts of a sum or difference that are common to the tops and bottoms of a fraction. In other words, (a+c)/(a+d) is generally not equal to c/d. For example, (7+2)/(7-2) should equal 9/5. But if we cancel out the 7's, we get the incorrect value of 2/-2 or -1. To cancel things out in fractions, each thing must be a factor of the top and bottom. For example, 9(2)/9(5)=18/45=2/5 or we could have just cancelled out the common factor of 9 in the beginning. This is a very common, and costly, mistake, so I would recommend spending some extra time making sure you are able to recognize when it's okay to cancel and when it's not.

Finally, if the problem HAD said n/2 instead of 2/n, it would simplify to (n+4)/(n-4)