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Find all A and B such that (2x + 7)/(x^2 - 8x + 15) = A/(x - 3) + B/(x - 5) for all x besides 3 and 5. Express your answer as an ordered pair in the form (A, B).

 Nov 12, 2020
 #1
avatar+111600 
+1

Find all A and B such that (2x + 7)/(x^2 - 8x + 15) = A/(x - 3) + B/(x - 5) for all x besides 3 and 5.

Express your answer as an ordered pair in the form (A, B).

 

\( \frac{(2x + 7)}{(x^2 - 8x + 15) }\\ = \frac{(2x + 7)}{(x-3)(x-5) }\\~\\ \frac{2x + 7}{(x-3)(x-5) }=\frac{A}{x-3}+\frac{B}{x-5}\\ \frac{2x + 7}{(x-3)(x-5) }=\frac{A(x-5)}{(x-3)(x-5) }+\frac{B(x-3)}{(x-3)(x-5) }\\ \)

 

 

You can finish it.    cool

 Nov 12, 2020
 #2
avatar+112375 
+1

Melody  has set up a partial fraction problem......thanks, Melody!!!

 

It may not be entirely clear  what to do after her last step

 

Note  that if we multiply through by the common denominator of  (x-3) (c-5)  we get that

 

2x+ 7 = A(x-5)  + B (x - 3)

 

2x + 7 = (A + B)x  -5A  - 3B 

 

This implies  that we have this system  :

 

A + B =  2

-5A - 3B = 7

 

Manipulating the first  we have that  B  = 2  - A

 

So....we  have  that

 

-5A - 3 ( 2 -A)  = 7

 

-5A  -6 + 3A  = 7

 

-2A  =  13

 

A = -13/2

 

And B  = 2 -  - (13/2)  =  17/2

 

 

cool cool cool

 Nov 12, 2020

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