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(2x-9)/((x^2)- x-6)=((A)/(x-3))+((B)/(x+2))
Find constants A and B that make the equation true.
Please explain how you got an answer! I originally tried using substitution, but the entire equation just became overly large, with none of the variables canceling out.
Added 3+ months ago:
Ignore the question mark in the question title -_-
(2x-9)/[(x+2)(x-3)] = [A(x+2)+B(x-3)]/[9x+2)(x-3)]
Therefore 2x-9 = Ax + 2A +Bx -3B; x not equal to 3 or -2 (division by zero)
2x = (A+B)x and -9 = 2A - 3B
A+B = 2. Therefore
3A + 3B = 6 and
2A - 3B = -9 Now adding gives
5A = -3; A = -3/5
-3/5 +B = 2 Therefore B = 10/5 + 3/5 = 13/5
Check A + B = -3/5 +13/5 = 10/5 = 2
