Justifying a solution for a system of equations in linear algebra I'm trying to understand how to solve systems of linear equations using matrices, however I'm faced with the following problem. Given the following system:
$$ 
\begin{cases}
x+2y+3z-3w=a \\
2x-5y-3x+12w=b \\
7x+y+8x+5w=c
\end{cases}
$$
What steps do I need to do to confirm and justify that it only has a single solution, which is:
$$
37a +13b = 9c
$$
Appreciate all hints! Thanks.

You can get your answer using simple elementary row operations: 
$$\left(
    \begin{matrix}
    1 & 2 & 3 & -3 & | & a\\
    2 & -5 & -3 & 12 & | & b \\ 
    7 & 1 & 8 & 5 & | & c \\
     \end{matrix}
\right)  >>{\text{$R2=R2-2R1, R3=R3-7R1$}}>>$$ $$\left(
    \begin{matrix}
    1 & 2 & 3 & -3 & | & a\\
    0 & -9 & -9 & 18 & | & b-2a \\ 
    0 & -13 & -13 & 26 & | & c-7a \\
     \end{matrix}
\right)>>{\text{$R3=R3+(- \frac{13} {9} R2)$}}>>$$
$$\left(
    \begin{matrix}
    1 & 2 & 3 & -3 & | & a\\
    0 & -9 & -9 & 18 & | & b-2a \\
    0 & 0 & 0 & 0 & | & c-7a + [- \frac{13} {9}(b-2a)]\\
     \end{matrix}
\right)$$ 
Now, notice that the last row of the matrix is a zero row, thus the system has a solution only if the right side of the equation is equal to zero. thus-
$$c-7a + [- \frac{13} {9}(b-2a)] =0$$
$$-\frac{37}{9}a-\frac{13}{9}b+c=0$$
$$-37a -13b +9c=0$$
$$37a+13b=9c$$
