<PRE><B>Question 95435</B>
Since the denominator of the 1st fraction is x^2+5
you need Ax+B as its numerator and C as the numerator of the 2nd fraction
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Then multiplying thru by (x^2+5)(x+9) you would get:
7 = (Ax+B)(x+9) + C(x^2+5)
7 = Ax^2+9Ax+Bx+9B+Cx^2+5C
7 = (A+C)x^2+(9A+B)x+(9B+5C)
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Let x=0, then 0A+9B+5C = 7
Let x=1, then 10A+10B+6C=7
Let x=-1, then -8A+8B+6C=7
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Solving that with a TI-83 Matrix facility  you get
A = -0.8139...
B = 0.73255...
C = 0.08139...
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That may or may not be what you are looking for but
you cannot arrive at an answer with A as the numerator over
x^2+5 and B as the numerator over x+9
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Cheers,
Stan H.


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