1. ## 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.

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

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

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

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

3. Hi thanks for helping.

You multiplied by the LCD?

4. Originally Posted by MrsPaiva
Hi thanks for helping.

You multiplied by the LCD?
I found a common denominator, and set the numerators equal.

5. So from here applied the distributive property and came up w/

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

Good so far?

6. 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.

7. My instructor wants us to show our work.

Thanks again

8. You can still show work using skeeter's method. This is what he was getting at:
Originally Posted by skeeter
$\displaystyle 5-x = A(x+1) + B(2x-1)$

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:
\displaystyle \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.

9. I see what you mean now. Okay, I misunderstood. I'll work out the problem and post my answer