How to solve systems of three equations? Either I forgot or never did learn to do it well. I need to solve the following system:
$$9a+3b+c=0$$
$$25a-5b+c=0$$
$$a-b+c=12$$
Google shows me this page with some instructions: http://www.jcoffman.com/Algebra2/ch3_6.htm, I decided to follow them.
The first one, is "add the first equation with the third one, this will eliminate an x-term". So I assume that, in my context, this will eliminate at least one term when I try it.
Adding the first with the third one:
$$9a+3b+c=0$$
$$a-b+c=12$$
I get:
$$10a-2b+2c$$
Aw... No term was removed. So something's not well.
Either my system is wrong or I am not following the instructions well. If you want to know where my system comes from, it is from the following question:

Determine the quadratic function such that $f(3) = 0$, $f(-5) = 0$ and
  $f(-1)=12$.

If I'm not mistaken, this involves solving the system I got above.
Can you tell me what did I do wrong following those instructions? I'm not really looking for the solution - instead, I'd prefer to understand how to do this.
 A: I will answer your question 

How to solve systems of three [linear] equations?

Don't necessarily take such instructions literally: what you refer to (adding the first equation to the third) was probably correct for the example used in the particular problem demonstrated. 
Essentially, solving a system of linear equations (aka Gaussian Elimination) is just like using elementary row operations in matrices, except you have the variables. But how you proceed depends on the coefficients:
$$9a+3b+c=0\tag{1}$$
$$25a-5b+c=0\tag{2}$$
$$a-b+c=12\tag{3}$$
To get you started:
Add $-9(3)$ to $(1)$: that eliminates $a$...
$\;\;-9a + 9 b - 9c = -108$
$+\; 9a + 3b + c = 0$
$= 0\; + 12b - 8c = -108$
$= 3b - 2c = -26\quad\quad\quad\quad (R_1)$
From here you can solve for for $b$ in terms of $c$, then back-substitute, etc.
Or, you can eliminate, say, $c$ altogether to solve for $b$:
For example, you can use $-25(\text{equation }\;3) + (\text{equation}\;2)$ to get a second equation without the $a$ variable...
$-25a + 25b - 25c = 300$
$+\; 25a - 5b + \;\;c = \;\;0$
$= 0\;\;20b - 24c = 300$
$= 5b - 6c = 75\quad\quad\quad\quad (R_2)$
Now continue the process using $3(R_1)+(R_2)$ to get rid of the $c$-term...
Then back substitute to solve for $b$, and then using $b, c$ solve for $a$.
A: I will answer your question

Determine the quadratic function such that  $f(3) = 0$, $f(-5) = 0$ and $f(-1) = 12$.

Any quadratic polynomial will have only two roots. From the question, you know that the only two roots are $3$ and $-5$. Hence, we have
$$f(x) = a(x-3)(x+5)$$
In addition, we are given thaat $f(-1) = 12$. This implies
$$a(-4)(4) = 12 \implies a = -\dfrac34$$
Hence, $$f(x) = -\dfrac34 (x-3)(x+5) = \dfrac34(3-x)(x+5)$$
To solve by your method, subtract equation $3$ from $1$ and $3$ from $2$ to eliminate $c$. If we do this, we then get that
$$8a+4b = -12$$
$$24a-4b = -12$$
Add the above two equations to get
$$32a = - 24 \implies a = - \dfrac34$$
We have $8a+4b = -12 \implies 4b = - 12 +6 = -6 \implies b = - \dfrac32$. Now that we have values for $a$ and $b$ plug it in any of the three original equations you had to get that $c = \dfrac{45}4$.
A: @Marvis' answer is the best way to go about getting that quadratic, but to answer "how do I solve this system of linear equations", the answer is: you can do other operations besides adding two equations to each other.  In particular, you can add a multiple of one equation to another.  So, if you wanted to eliminate $a$ from the first equation, you could multiply the third equation by $-9$ then add it to the first:
$$
\begin{align*}
-9(a-b+c)&=-9\cdot 12\\
+ (9a+3b+c)&=0\\
\Rightarrow -9a+9a+9b+3b-9c+c&=-9\cdot 12
\end{align*}
$$
The first equation would then read: 
$$
12b-8c=-108
$$
From here you could put $b$ in terms of $c$, and substitute, etc.  This process is called Gaussian elimination.  There are lots of good YouTube videos on the subject as well.
A: Another solution is using Matrix. You are three variable, $a $, $b$ and $c$.
$$
9a+3b+c=0\\
25a−5b+c=0\\
a−b+c=12
$$
These three equations are equivalent to Matrix Production in form $A.X=b$, $A$ like this:
$$
\begin{bmatrix}
9 & 3 & 1\\
25 & -5 & 1 \\
1 & -1 & 1
\end{bmatrix}
\begin{bmatrix}
a\\
b\\
c
\end{bmatrix}
=
\begin{bmatrix}
0\\
0\\
12
\end{bmatrix}
$$ 
that
$$
A = \begin{bmatrix}
9 & 3 & 1\\
25 & -5 & 1 \\
1 & -1 & 1
\end{bmatrix}, 
X = 
\begin{bmatrix}
a\\
b\\
c
\end{bmatrix},
b =
\begin{bmatrix}
0\\
0\\
12
\end{bmatrix}
$$
now if $det(A)\neq 0$ (Determinant of matrix $A$), the system has a unique solution given by:
$$
X=A^{-1}.b
$$
because:
$$
\begin{align}
& A.X=b \\
&\Rightarrow A^{-1}.A.X=A^{-1}.b\\
 &\Rightarrow I.X=A^{-1}.b\\
 &\Rightarrow X=A^{-1}.b
\end{align}
$$
if you calculate Inversion of $A$:
$$
A^{-1} =
\begin{bmatrix}
1/32 & 1/32 & -1/16 \\
3/16 & -1/16 & -1/8 \\
5/32 & -3/32 & 15/16
\end{bmatrix}
$$
so
$$
X = 
\begin{bmatrix}
1/32 & 1/32 & -1/16 \\
3/16 & -1/16 & -1/8 \\
5/32 & -3/32 & 15/16
\end{bmatrix}
\begin{bmatrix}
0\\
0\\
12
\end{bmatrix}
$$
Another way is Gaussian Elimination.
