Partial fraction of $\frac{2s+12}{ (s^2 + 5s + 6)(s+1)} $ then inverse transform it 
Find the inverse Laplace transform of
  $$\mathcal{L}^{-1}\left(\frac{2s+12}{ (s^2 + 5s + 6)(s+1)}\right)$$

I recognise I need to use partial fractions to solve it and that is where I got stuck. Here’s my working, 
After factoring the denominator, I got a case of non repeating linear factors
$\dfrac{2s+12}{ (s+2)(s+3)(s+1)} = \dfrac{A}{s+2} + \dfrac{B}{s+3} + \dfrac{C}{s+1} $ 
$2s + 12 = A (s+3)(s+1) + B (s+2)(s+1) + C (s+2)(s+3) $ 
$2s + 12 = (As^2 + Bs^2 + Cs^2) + (4As + 3Bs + 5Cs) + (3A + 2B+6C) $ 
So...
$A + B + C = 0$
$ 4A + 3B + 5C = 2$
$3A + 2B + 6C = 12$ 
how do I solve this complicated equations ? This is where i got stuck and cannot continue. 
 A: You can use the methods associated with the "Partial fraction decomposition" (see https://en.wikipedia.org/wiki/Partial_fraction_decomposition). After writing :
$$\frac{2s+12}{ (s+2)(s+3)(s+1)} = \frac{A}{s+2} + \frac{B}{s+3} + \frac{C}{s+1} $$  
You multiply and the right and left hand side by $(s+2)$, which gives : 
$$\frac{2s+12}{(s+3)(s+1)} =A + \frac{B(s+2)}{s+3} + \frac{C(s+2)}{s+1} $$ 
Taking $s = -2$, you find :
$$  \frac{2(-2)+12}{(-2+3)(-2+1)} = \frac{8}{-1} = -8 = A  $$
You can find $B$ and $C$ using the same method (which is usually easier than solving the system).
A: There is a shortcut way here, in this simple case. Let to mutiply the relation to $\color{red}{s+1}$, then
$$\frac{2s+12}{ (s+2)(s+3)(s+1)} \color{red}{(s+1)}= \frac{A\color{red}{(s+1)}}{s+2} + \frac{B\color{red}{(s+1)}}{s+3} + \frac{C\color{red}{(s+1)}}{s+1}$$
or
$$\frac{2s+12}{ (s+2)(s+3)} = \frac{A\color{red}{(s+1)}}{s+2} + \frac{B\color{red}{(s+1)}}{s+3} + C$$
now set $s=-1$ then $C=\dfrac{10}{2}$. 
Can you proceed?
A: $A + B + C = 0\quad[1]$
$ 4A + 3B + 5C = 2\quad[2]$
$3A + 2B + 6C = 12\quad[3]$
$ $
$[2]-3[1]$
$A+2C=2\quad[4]$
$ $
$[3]-2[1]$
$A+4C=12\quad[5]$
$ $
$[5]-[4]$
$2C=10$
$ $
$\therefore C=5$
$\therefore A=-8$
$\therefore B=3$
