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 April 23rd, 2012, 03:26 AM #1 Member   Joined: Feb 2012 Posts: 97 Thanks: 0 Add fraction $\frac {2}{3x^2-4x-7}$ + $\frac {1}{3x^2-19x+28}$ $\frac {2}{(3x-7)(x+1)}$ $\frac {1}{(3x-7)(x-4)}$ $\frac {2}{(x-4)+1(x-1)}\frac {1}{(x-4)(3x-7)(x-1)}$ ; $\frac {3x-9}{(x-4)(3x-7)(x-1)}$ Is this right?
 April 23rd, 2012, 04:08 AM #2 Member   Joined: Nov 2011 Posts: 45 Thanks: 0 Re: Add fraction $\frac {2}{3x^2-4x-7}$ + $\frac {1}{3x^2-19x+28}$ $\frac {2(x-4)+(x+1)}{(3x-7)(x-4)(x+1)}$
April 23rd, 2012, 05:13 AM   #3
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Quote:
 Originally Posted by ray $\frac {2}{3x^2-4x-7}$ + $\frac {1}{3x^2-19x+28}$ $\frac {2(x-4)+(x+1)}{(3x-7)(x-4)(x+1)}$
Agree Ray; can be further simplified to:
1 / [(x - 4)(x + 1)]

Did you forget that a/b + c/d = (ad + bc) / (bd) ?

 April 23rd, 2012, 06:29 PM #4 Member   Joined: Feb 2012 Posts: 97 Thanks: 0 Re: Add fraction $\frac {2}{(3x-7)(x+1)}$ and that's it?
 April 23rd, 2012, 06:47 PM #5 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,935 Thanks: 1129 Math Focus: Elementary mathematics and beyond Re: Add fraction $\frac{2}{3x^2\,-\,4x\,-\,7}\,+\,\frac{1}{3x^2\,-\,19x\,+\,28}\,=\,\frac{2}{(3x\,-\,7)(x\,+\,1)}\,+\,\frac{1}{(3x\,-\,7)(x\,-\,4)} \\ =\,\frac{2(x\,-\,4)}{(3x\,-\,7)(x\,-\,4)(x\,+\,1)}\,+\,\frac{x\,+\,1}{(3x\,-\,7)(x\,-\,4)(x\,+\,1)} \\ =\,\frac{2x\,-\,8}{(3x\,-\,7)(x\,-\,4)(x\,+\,1)}\,+\,\frac{x\,+\,1}{(3x\,-\,7)(x\,-\,4)(x\,+\,1)} \\ =\,\frac{2x\,-\,8\,+\,x\,+\,1}{(3x\,-\,7)(x\,-\,4)(x\,+\,1)} =\,\frac{3x\,-\,7}{(3x\,-\,7)(x\,-\,4)(x\,+\,1)} \\ =\,\frac{1}{(x\,-\,4)(x\,+\,1)}$
 April 23rd, 2012, 08:02 PM #6 Member   Joined: Feb 2012 Posts: 97 Thanks: 0 Re: Add fraction Thanks. $\frac {(w-4)(w+5)}{4(w-1)(w+1)}$ ... right path?
 April 24th, 2012, 04:31 AM #7 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Add fraction You have factored the numerator correctly, but not the denominator: $\frac{w^2+w-20}{64-4w^2}=\frac{(w+5)(w-4)}{4(4+w)(4-w)}=-\frac{(w+5)(4-w)}{4(4+w)(4-w)}=-\frac{w+5}{4(w+4)}$

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