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 lalex0710 September 24th, 2012 12:47 PM

Rational expression constants

Find, Algebraically, the constants A and B such that 7x-6 / x^2-x-6 = A/x+2 + B/x-3

 greg1313 September 24th, 2012 01:20 PM

Re: Rational expression constants

Will you post your attempt and/or any ideas you have?

 lalex0710 September 24th, 2012 01:43 PM

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

 greg1313 September 24th, 2012 01:47 PM

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?

 lalex0710 September 24th, 2012 02:11 PM

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

 greg1313 September 24th, 2012 02:14 PM

Re: Rational expression constants

$7x\,-\,6\,=\,A(x\,-\,3)\,+\,B(x\,+\,2) \\
7(3)\,-\,6\,=\,A(3\,-\,3)\,+\,B(3\,+\,2) \\
15\,=\,5B \\
B\,=\,3$

 lalex0710 September 24th, 2012 02:18 PM

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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