Partial fraction in two variable problem

Question

How to write partial fraction of $$\frac{12m-n-3mn+7}{5m-2n-2mn+5}$$
I just write first and second denominator: $5-2n$ and $m+1$.

Answer 1

HINT:

$$\frac{12m-n-3mn+7}{5m-2n-2mn+5} = \frac{12m-n-3mn+7}{(m+1)(5-2n)}$$ $$= \frac{A}{m+1}+\frac{B}{5-2n}$$ $$= \frac{A(m+1)+B(5-2n)}{(m+1)(5-2n)}$$ where $A,B$ are constants to be determined.

So we can write that $$A(m+1)+B(5-2n)=12m-n-3mn+7$$

Can you proceed now?

Answer 2

The deg on top and bottom are the same. Before we do the partial fraction thing we should do division I believe. So the division part: $\frac{12m-n-3mn+7}{5m-2n-2mn+5}=\frac{\frac{3}{2}(-2mn+5-2n+5m)}{5m-2n-2mn+5}+\frac{-\frac{1}{2}+2n+\frac{9}{2}m}{5m-2n-2mn+5}$ ... Now preparing for the partial fraction part... $-2mn+5-2n+5m=-2mn+2n+5+5m=-2n(m+1)+5(m+1)=(m+1)(5-2n)$... So we have $\frac{\frac{-1}{2}+2n+\frac{9}{2}m}{(m+1)(5-2n)}=\frac{A}{m+1}+\frac{B}{5-2n}$ giving us $\frac{-1}{2}+2n+\frac{9}{2}m=A(5-2n)+B(m+1)$ Expanding... $\frac{-1}{2}+2n+\frac{9}{2}m=-2An+5A+Bm+B=-2An+Bm+(5A+B)$ Now compare both sides: $2=-2A ; B=\frac{9}{2} ;5A+B=\frac{-1}{2}$ You should end up with $\frac{3}{2}+\frac{-1}{m+1}+\frac{\frac{9}{2}}{5-2n}$