Partial fractions- Equating coefficients This is a partial fraction question. Kindly could you assist me with finding the values for A, B and C as I'm not very clear on how to go about it. The question is as follows;
$$\frac{3-x}{(x^2+3)(x+3)}$$
And we want A, B, C so that  $$\frac{3-x}{(x^2+3)(x+3)} = \frac{Ax+B}{x^2 + 3} + \frac C{x+3}$$
I know that $A+B= 3$, $3B+C= -1$ and $3A+3C= 0$
The problem I'm facing here is with equating coefficients. I also know that after equating coefficients, the final values you get for A,B and C are as follows;
\begin{align}
A&= -1/2\\
B&= 1/2\\
C&= 1/2
\end{align}
But how do you solve with equating coefficients? Can you provide me step-by-step working for this please. I'm not facing any difficulty with partial fractions but just equating coefficients part in this case. Please help.
Many thanks.

It looks like you already have equated coefficients:
$$\frac{3-x}{(x^2+3)(x+3)}=\frac{Ax+B}{x^2+3}+\frac C{x+3}$$
$$\iff \frac{\color{blue}{(Ax+B)(x+3)+C(x^2+3)}}{(x^2+3)(x+3)}= \frac{\color{blue}{3-x}}{(x^2+3)(x+3)}$$
Equating the numerators:
$$(Ax+B)(x+3)+C(x^2+3) = 0\cdot x^2 -x + 3\tag{1}$$
Expanding the LHS of equation $(1)$, gathering like terms:
$$Ax^2 + (B + 3A)x + 3B + Cx^2 + 3C = 0x^2 - x + 3$$
$$\iff \color{blue}{\bf (A + C)}x^2 +\color{red}{\bf (3A + B)}x + \color{green}{\bf (3B + 3C)} = \color{blue}{\bf 0}x^2 + \color{red}{\bf (-1)}x + \color{green}{\bf 3}\tag{2}$$
Match up (color coded above) coefficients on LHS with those on RHS of $(2)$:
$$\iff \color{blue }{\bf A + C = 0}, \quad \color{red}{\bf 3A + B = -1}, \quad \color{green}{\bf 3(B + C) = 3 \iff B+C = 1}$$
Indeed, this gives us a system of $\bf 3$ equations in $\bf 3$ unknowns, from which we can solve for the "unknowns" $A, B, \;\text{and}\; C$.
$$\begin{align} A + C & = 0 \tag{i}\\  3A + B & = -1 \tag{ii}\\ B+ C & = 1\tag{iii}\end{align}$$
Subtract $(iii)$ from $(i)$: $A - B = -1\tag{iv}$
Adding $(iv)$ to $(ii)$ gives us $4A = -2 \iff A = -\dfrac 12\tag{A}$
From $(i)$: $A = -\dfrac 12 \implies C = \dfrac 12\tag{C}$
From $(iii)$: $C = \dfrac 12 \implies B = \dfrac 12\tag{B}$
Therefore, we have the following function, replacing coefficients A, B, C with their found values:
$$\frac{3-x}{(x^2+3)(x+3)}=\frac{Ax+B}{x^2+3}+\frac C{x+3} $$ $$= \frac{-\frac12\cdot x+\frac 12}{x^2+3}+\frac {\frac 12}{x+3}=\dfrac 12\left(\frac{1-x}{x^2 + 3}\right) + \dfrac 12\left(\frac 1{x+3}\right)$$
