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.

Step 1: Write $$\frac{3-x}{(x^2+3)(x+3)}=\frac{Ax+B}{x^2+3}+\frac{C}{x+3}$$
Step 2: Expand the RHS to get
$$\frac{Ax+B}{x^2+3}+\frac{C}{x+3}=\frac{(Ax+B)(x+3)+C(x^2+3)}{(x^2+3)(x+3)}=\frac{(A+C)x^2+(3A+B)x+3(B+C)}{(x^2+3)(x+3)}$$
Step 3: Compare the coefficients.
$$x^2:A+C=0,\quad x:(3A+B)=-1,\quad\text{constant}: 3(B+C)=3$$
Step 4: Solve simultaneous equations.
