I am trying to find a partial fraction decomposition for the following:
$$\frac{1}{(-\alpha xi+4y)(\alpha xi + 2y)}$$
where $\alpha\in \mathbb{R}$. I am understanding that I could write this expression as a partial fraction decomposition as:
$$\frac{1}{(-\alpha xi+4y)(\alpha xi + 2y)}=\frac{A}{-\alpha xi+4y}+\frac{B}{\alpha xi+2y}~~(*)$$
but considering that this yields
$$1=A(\alpha xi+2y)+B(-\alpha xi+4y)\implies \\ 1=A\alpha xi+2Ay-B\alpha xi+4By \implies \\1=(A - B)\alpha xi+(2A+4B)y$$
I cannot understand what has gone wrong here, since this does not seem to yield any reasonable answers, at least as far as I understand. Has something gone wrong in my decomposition in $(*)$?
The following analysis assumes that $0 \neq x,y,a$.
As Ross Millikan's comment indicates, I am also assuming that $x,y \in \Bbb{R}.$
$\displaystyle \frac{1}{(-axi + 4y)(axi + 2y)} = \frac{A}{-axi + 4y} + \frac{B}{axi + 2y} \implies$
$$A(axi + 2y) + B(-axi + 4y) = 1.$$
This implies that $A - B = 0$ and $(2y)(A + 2B) = 1.$
So, you have two linear equations in the $2$ unknowns, $A$ and $B$.