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 September 24th, 2012, 12:47 PM #1 Newbie   Joined: Aug 2012 Posts: 16 Thanks: 0 Rational expression constants Find, Algebraically, the constants A and B such that 7x-6 / x^2-x-6 = A/x+2 + B/x-3
 September 24th, 2012, 01:20 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,968 Thanks: 1152 Math Focus: Elementary mathematics and beyond Re: Rational expression constants Will you post your attempt and/or any ideas you have?
 September 24th, 2012, 01:43 PM #3 Newbie   Joined: Aug 2012 Posts: 16 Thanks: 0 Re: Rational expression constants yea sure, here they are: ive tried using (x+2)(x-3) as the denominator and took it out from all 3 then, ax + 3a + bx + 2b = 7x-6 ive tried substituting a and couldnt go any further than this
 September 24th, 2012, 01:47 PM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,968 Thanks: 1152 Math Focus: Elementary mathematics and beyond Re: Rational expression constants $\frac{7x\,-\,6}{(x\,-\,3)(x\,+\,2)}\,=\,\frac{A}{x\,+\,2}\,+\,\frac{B}{ x\,-\,3}\,\Leftrightarrow\,7x\,-\,6\,=\,A(x\,-\,3)\,+\,B(x\,+\,2)$ Now set x = 3 and you get B = 3. Do you see how?
 September 24th, 2012, 02:11 PM #5 Newbie   Joined: Aug 2012 Posts: 16 Thanks: 0 Re: Rational expression constants I've tried following your suggestions but couldn't quite get the 3 for B, however what I got was 7x-6 = B by eliminating all the like factors and can you also elaborate on the rules of plugging in the 3 as x. Thanks
 September 24th, 2012, 02:14 PM #6 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,968 Thanks: 1152 Math Focus: Elementary mathematics and beyond Re: Rational expression constants $7x\,-\,6\,=\,A(x\,-\,3)\,+\,B(x\,+\,2) \\ 7(3)\,-\,6\,=\,A(3\,-\,3)\,+\,B(3\,+\,2) \\ 15\,=\,5B \\ B\,=\,3$
 September 24th, 2012, 02:18 PM #7 Newbie   Joined: Aug 2012 Posts: 16 Thanks: 0 Re: Rational expression constants alright I see, I just made a small error and forgot to put 5 infront of B. Anyways, thanks for clearing this up.

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