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October 1st, 2007, 12:52 PM  #1 
Newbie Joined: Oct 2007 Posts: 3 Thanks: 0  Yes, this may be an easy question, but im stumped
ok, so i've just entered college after taking a year off after high school. So I have COMPLETELY lost all my math skills.....so if you could run me throught the process of how to complete this question, it should clear up some of the others. Thanks! Simplify: (3/x) + [1/(x^2 + x)]  (1/x+1)  (1/x1) The answer in the book is : (3x+4)(x1)  2x 
October 1st, 2007, 03:35 PM  #2 
Senior Member Joined: Sep 2007 From: USA Posts: 349 Thanks: 67 Math Focus: Calculus 
For the denominator, do you mean 1/(x + 1)  1/(x  1) or 1/x + 1  1/x + 1?

October 1st, 2007, 03:37 PM  #3 
Newbie Joined: Oct 2007 Posts: 3 Thanks: 0 
1/(x + 1)  1/(x  1) sorry for the confusion 
October 1st, 2007, 04:02 PM  #4 
Senior Member Joined: Sep 2007 From: USA Posts: 349 Thanks: 67 Math Focus: Calculus 
What I would do is multiply the numerator and the denominator by (x+1)(x1), which would eliminate the fractions in the denominator. Then you can simplify the numerator.

October 1st, 2007, 04:15 PM  #5 
Newbie Joined: Oct 2007 Posts: 3 Thanks: 0 
i still dont understand how you can just multiply both denominator and numerator. both the numerator and denominator are fractions, so i dont understand how the process works.

October 1st, 2007, 04:32 PM  #6 
Senior Member Joined: Sep 2007 From: USA Posts: 349 Thanks: 67 Math Focus: Calculus 
Doing this is multiplying by 1, because (x+1)(x1)/[(x+1)(x1)]=1. For the numerator, you distribute (x+1)(x1) across the terms. You do the same for the denominator. [3/x + 1/(x(x+1))](x+1)(x1)  [1/(x+1)  1/(x1)](x+1)(x1) 3(x+1)(x1)/x + (x1)/x  (x1)  (x+1) (3x^2+x4)  2x (3x+4)(x1)  2x 

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