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Find constants A and B such that \(\frac{x + 7}{x^2 - x - 2} = \frac{A}{x - 2} + \frac{B}{x + 1}\) for all x such that \(x\neq -1\) and \(x\neq 2\). Give your answer as the ordered pair (A,B).

Lightning  Aug 18, 2018
 #1
avatar+93866 
+3

\frac{x + 7}{x^2 - x - 2} = \frac{A}{x - 2} + \frac{B}{x + 1}

 

\(\frac{x + 7}{x^2 - x - 2} = \frac{A}{x - 2} + \frac{B}{x + 1}\\ A(x+1)+B(x-2)=x+7\\ Ax+Bx=x \qquad A-2B=7\\ A+B=1 \qquad A=7+2b\\ 7+2B+B=1\\ 7+3B=1\\ 3B=-6\\ B=-2\\ A=3\\~\\ check\\ 3(x+1)-2(x-2)=x+7 \quad great \)

 

answer   (3,-2)

Melody  Aug 18, 2018
 #2
avatar+27128 
+2

Sorry Melody, I didn't realise you were working on this.

 

This question seems to have appeared twice!  I answered the other one at https://web2.0calc.com/questions/help-please_18822#r1.

Alan  Aug 18, 2018
edited by Alan  Aug 18, 2018
 #3
avatar+93866 
+1

That is not a problem Alan   laugh

 

Maybe 2 answers are better than one and it certainly is not your fault if Lightning double posted.

Melody  Aug 18, 2018
 #4
avatar+803 
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I didn't double post this one. I would have realized the answer was already answered. Plus, I have a work from 9pm to 3am. So it coudn't be me.

Lightning  Aug 18, 2018
 #5
avatar+93866 
0

That is ok Lightning. The oether was someone else in your class I expect.

 

It would be nice if you added a 'thank you' for Alan and me for answering you though.   laugh

Melody  Aug 19, 2018

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