# Partial fraction decomposition

• Sep 10th 2007, 03:52 PM
OnMyWayToBeAMathProffesor
Partial fraction decomposition
hello,

i have a problem i am stuck on. i have figured it out till the very end but can not find the answer.here it is.

The Problem : Find the partial fraction decomposition.
(-x+10)/(x^2+x-12)

here is what i got from that problem.

(-x+10)/(x+4)(x-3) = (A/(x+4))+(B/(x-3))

(-x+10)=A(x-3)+B(x+4)

(-x+10)=Ax-3A+Bx+B4

-1=A+B

10= (-3a)+(b4)

What is A and B so (A/(x+4))+(B/(x-3)) = (-x+10)/(x+4)(x-3)

Thank you.
• Sep 10th 2007, 05:37 PM
Jhevon
Quote:

Originally Posted by OnMyWayToBeAMathProffesor
hello,

i have a problem i am stuck on. i have figured it out till the very end but can not find the answer.here it is.

The Problem : Find the partial fraction decomposition.
(-x+10)/(x^2+x-12)

here is what i got from that problem.

(-x+10)/(x+4)(x-3) = (A/(x+4))+(B/(x-3))

(-x+10)=A(x-3)+B(x+4)

(-x+10)=Ax-3A+Bx+B4

-1=A+B

10= (-3a)+(b4)

What is A and B so (A/(x+4))+(B/(x-3)) = (-x+10)/(x+4)(x-3)

Thank you.

so i guess you are having trouble solving the simultaneous equations?

here's another approach

\$\displaystyle -x + 10 = A(x - 3) + B(x + 4)\$

Let \$\displaystyle x = 3 \implies -3 + 10 = 7B\$
Let \$\displaystyle x = -4 \implies 4 + 10 = -7A\$

there you have your A and B

the method you were attempting was the equating coefficients method, i don't see the problem you have, the only thing left for you to do is to solve the simultaneous equations
• Sep 11th 2007, 02:48 PM
OnMyWayToBeAMathProffesor