# partial-fraction-decomp question

#### MrsPaiva

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.

#### skeeter

MHF Helper
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

#### MrsPaiva

Hi thanks for helping.

You multiplied by the LCD?

#### skeeter

MHF Helper
Hi thanks for helping.

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

#### MrsPaiva

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

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

Good so far?

#### skeeter

MHF Helper
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.

#### MrsPaiva

My instructor wants us to show our work.

Thanks again

#### eumyang

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

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.

#### MrsPaiva

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