Just don't see how to get started with this one

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I don't see how to solve a problem like this one:

If x is different from 3, then (3x(x-3) -x +3)/(x-3) = ...

A : X+1
B : X-3
C : 3x +1
D : 3x -1
E : 3X +3

Any help welcome! Thanks!

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by Pugalenthi » Fri May 11, 2012 12:15 pm
The answer is D
3x(x-3) -x +3/x-3 could be written as
3x(x-3) -1(x-3)/x-3
(3x-1)(x-3)/x-3
3x -1

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by GMATBeater77 » Fri May 11, 2012 12:45 pm
Thanks, that's totally correct.
I understand your approach however I wouldn't have come to this solution myself.
Is there some kind of strategy to approach these kind of problems.

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by Bill@VeritasPrep » Fri May 11, 2012 4:03 pm
GMATBeater77 wrote:Thanks, that's totally correct.
I understand your approach however I wouldn't have come to this solution myself.
Is there some kind of strategy to approach these kind of problems.
The biggest thing is to look for an opportunity to factor out a common term. In this case, the fact that we have (x-3) in the numerator and the denominator should point us in the right direction.
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by GMATGuruNY » Fri May 11, 2012 7:00 pm
GMATBeater77 wrote:I don't see how to solve a problem like this one:

If x is different from 3, then (3x(x-3) -x +3)/(x-3) = ...

A : X+1
B : X-3
C : 3x +1
D : 3x -1
E : 3X +3

Any help welcome! Thanks!
Let x=0.
Then (3x(x-3) -x +3)/(x-3) = (0-0+3)/(0-3) = -1. This is our target.

Now we plug x=0 into the answers to see which yields our target of -1.
Only D works:
3x-1 = 3(0)-1 = -1.

The correct answer is D.
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