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).
+118608 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. 
+128408 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
