When (19x - 8)/(2x^2 - x - 21) is decompossed into partial fractions, what is the sum of the numerators when each fraction is reduced to lowest terms?
2006-09-23 04:21:24 · 3 answers · asked by Sasha 2 in Science & Mathematics ➔ Mathematics
(19x - 8) / (2x^2 - x - 21) =(19x-8) / ((2x-7)(x+3)) = A/(2x-7) + B/(x+3)
= (A(x+3) +B(2x-7)) / ((2x-7)(x+3))
= ((A+2B)x + (3A-7B)) / (2x^2 - x - 21)
So we need 19=A+2B and -8=3A-7B
so A=19-2B
so -8=3(19-2B)-7B=57-13B so B=(57+8)/13 = 65/13 = 5
and A=19-2B = 19-10 = 9
So the partial fraction decomposition is
(19x-8)/(2x^2 - x -21) = 9/(2x-7) + 5/(x+3)
So I guess the answer is 9+5 = 13, if I understand the second part of the question correctly.
2006-09-23 08:45:15 · answer #1 · answered by vinzklorthos 2 · 0⤊ 0⤋
14
2006-09-23 11:43:26 · answer #2 · answered by Greg G 5 · 0⤊ 0⤋
You have got be kidding... and you want extra credit for this ??? ROFLMAO!!!!
2006-09-23 11:29:30 · answer #3 · answered by WizD 3 · 0⤊ 0⤋