How to simplify $\frac{2mn+4m+3n+2}{2m-mn+2-n}$ ?

Question

I was dealing with this problem:

Find $\inf A$ and $\sup A$ if

$$A=\{\frac{2mn+4m+3n+2}{2m-mn+2-n}: m,n\in\Bbb N\}$$

I have to separate this set into 2 sets to make things easier, like this:

$A_1=\{-\frac{2m+3}{m+1}:m\in \Bbb N\}$ and $A_2=\{\frac{8}{2-n}:n\in\Bbb N\}$

So $A=A_1+A_2$. Now I would know how to find $\inf$ and $\sup$, but my question is how did someone come up with $$\frac{2mn+4m+3n+2}{2m-mn+2-n}=-\frac{2m+3}{m+1}+\frac{8}{2-n}$$ I understand that $2m-mn+2-n=(m+1)(2-n)$ but what do you do in the numerator to get something that will cancel out?

Answer 1

$2nm+4m+3n+2=2mn-4m+3n-6+8m+8=-(2m+3)(2-n)+8(m+1)$ Now when u separate the two you get

$$\frac{-(2m+3)(2-n)}{(m+1)(2-n)}+\frac{8(m+1)}{(m+1)(2-n)}$$

from there it's obvious how the last expression is achieved.

Answer 2

the denominator can written as $$m(2-n)+2-n=(2-n)(m+1)$$ for the numerator we have $$2m(n+2)+n+2+2n=(n+2)(2m+1)+2n$$ other possibilities for the fraction are $$-{\frac {2\,m+7}{m+1}}+{\frac {-16-4\,m}{ \left( n-2 \right) \left( m +1 \right) }} $$ or $$-2\,{\frac {n}{n-2}}+{\frac {-5\,n-2}{ \left( n-2 \right) \left( m+1 \right) }} $$

Answer 3

$$\frac{2mn+4m+3n+2}{2m-mn+2-n}=\frac{A}{m+1}+\frac{B}{2-n}$$

$$(2mn+4m+3n+2)=A.(2-n)+B.(m+1)$$

Put m = -1 and calculate A.

Then put n = 2 and calculate B.

You have terms for numerator.