# partial-fraction-decomp question

• Jul 20th 2010, 12:47 PM
MrsPaiva
partial-fraction-decomp question
I'm having such a hard time solving the problem below

5-x
_____
2xsquared + x -1

Can you help point me in the right direction?

Thank you very much.
• Jul 20th 2010, 01:15 PM
skeeter
Quote:

Originally Posted by MrsPaiva
I'm having such a hard time solving the problem below

5-x
_____
2xsquared + x -1

$\displaystyle \frac{5-x}{(2x-1)(x+1)} = \frac{A}{2x-1} + \frac{B}{x+1}$

$5-x = A(x+1) + B(2x-1)$

choose two "strategic" values for x to help you find A and B
• Jul 20th 2010, 03:19 PM
MrsPaiva
Hi thanks for helping.

You multiplied by the LCD?
• Jul 20th 2010, 03:32 PM
skeeter
Quote:

Originally Posted by MrsPaiva
Hi thanks for helping.

You multiplied by the LCD?

I found a common denominator, and set the numerators equal.
• Jul 20th 2010, 04:10 PM
MrsPaiva
So from here applied the distributive property and came up w/

5-X = AX + 1A +2BX -1B

Good so far?
• Jul 20th 2010, 04:29 PM
skeeter
Quote:

Originally Posted by MrsPaiva
So from here applied the distributive property and came up w/

5-X = AX + 1A +2BX -1B

Good so far?

you can do it this way (equating coefficients and solving a system), although the method I recommended in my initial post is much easier, imho.
• Jul 20th 2010, 04:43 PM
MrsPaiva
My instructor wants us to show our work.

Thanks again
• Jul 20th 2010, 06:32 PM
eumyang
You can still show work using skeeter's method. This is what he was getting at:
Quote:

Originally Posted by skeeter
$5-x = A(x+1) + B(2x-1)$

choose two "strategic" values for x to help you find A and B

Choose a value of x such that the A is eliminated. So let x = -1:
\begin{aligned}
5 + 1 &= A(-1 + 1) + B(2(-1) - 1) \\
6 &= A(0) + B(-3) \\
6 &= -3B \\
B &= -2
\end{aligned}

See what happens? You'll just have one variable B, that you can solve for it easily.

Now choose a value of x so that B is eliminated.
• Jul 20th 2010, 06:39 PM
MrsPaiva
I see what you mean now. (Doh) Okay, I misunderstood. I'll work out the problem and post my answer