SOLUTION: I need help solving this complex fraction please? 5/(2x^2-18) - 2/(x^2-7x+12) all over 2/(x^2-x-12) - 4/(3x^2-27) Thank you

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Question 882289: I need help solving this complex fraction please?
5/(2x^2-18) - 2/(x^2-7x+12) all over 2/(x^2-x-12) - 4/(3x^2-27)
Thank you
Found 2 solutions by DrBeeee, Edwin McCravy:
Answer by DrBeeee(684)   (Show Source): You can put this solution on YOUR website!
I'm very slow at typing etc so I wont show all the steps.
The most important step is to FACTOR ALL FOUR denominators of the given expression;
(1) 2x^2 - 18 = 2*(x-3)*(x+3)
(2) x^2 -7x +12 = (x-3)*(x-4)
(3) x^2-x-12 = (x-4)*(x+3)
(4) 3x^2-27 = 3*(x-3)*(x+3)
Then factor out a common factor of the numerator
(5) 1/(x-3) and from the denominator
(6) 2/(x+3)
Add the two similar fraction in the numerator to get
(7) (x-32)/(2*(x+3)*(x-4)), and add the denominator fractions to get
(8) (x-1)/(3*(x-4)*(x-3))
At this point you have a few common factors that cancel, namely,
(9) (x-3),(x+3) and (x-4) leaving your answer
(10) 3/4*((x-32)/(x-1))
Note that you should include any values of x that render the expression undedined (divide by zero), they are x not equal to {1,-3,+3,4}
Answer by Edwin McCravy(20060)   (Show Source): You can put this solution on YOUR website!
Here is a way that shows even more detail:



As the good doctor above says, factor all the four denominators

Factor: 
Factor: 
Factor: 
Factor: 



The LCD of all those denominators is 

Put it over 1: 

and multiply every term in the numerator and denominator by it:



Then you cancel as much as you can:









You can leave it like that or else you can factor out a common
factor in the numerator and denominator:



Edwin
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