Simultaneous Equations - System of Equations Question and answer from an exam paper, how on earth are you supposed to work this out from the beginning. I can see how the answer works, but for a similar question how do I figure out what to times/add by what to make the two equal.
Question:
Find the value of a which allows
solutions to the system of equations
4x − y + 2z = 7
x + y − 3z = −3
2x − 3y + 8z = a

Solution:
4x − y + 2z = 7 [1]
x + y − 3z = −3 [2]
2x − 3y + 8z = a [3]
[1] + [2], 5x − z = 4
3 × [2] + [3], 5x − z = −9 + a
Equations are consistent if 4 = −9 + a i.e. a = 13


Another way of asking the question; is there a logical method for working these types of questions out.
 A: 
Question:  Find the value of a which allows solutions to the system of equations:
$4x −\; y + 2z = \;\; 7 \qquad\; [1]$
$\;\; x + \;y − 3z = −3 \qquad [2]$
$2x − 3y + 8z = \; a \qquad\;  [3]$

Solution:
First of all, take a look at equations $[1],[2],[3]$ and note any similarities between the coefficients of the $x$'s, $y$'s, and $z$'s. The goal is to reduce the system of equations in such a way that we can solve for a single variable, or, in this case, we need only reduce the system to two equations in $x$ and $z$.
I. Note that in the 1st and 2nd equation, the coefficients of the $y$-terms are, respectively, $-1$ and $1$.  So simply by adding equations $[1]$ and $[2]$, we can eliminate $y$.
II. That said, it would then make the most sense to eliminate the $y$-variable in the third equation, as well; we can do so my multiplying equation $[2]$ by $3$, then adding the modified equation $[2]$ to $[3]$.
We thus end up with two equations which resolve the problem nicely.  More details:
$[1] + [2] \to 5x − z = 4\qquad$ Add eq. $[1]$; the $y$-terms cancel out, leaving $5x - z = 4$: 
$\; 4x-y+2z = 7$
$\underline{+ x + y - 3z = -3}$ 
$\;5x+0y-z =4 \implies 5x - z =4\qquad$ [I]
$3 \times [2] + [3] \to 5x-z=-9\quad$  Multiply $[2]$ by $3$: $\quad 3(x+y-3z=-3) = 3x +3y - 9z=-9$ 
*Note: Take this last equation [$3x +3y - 9z=-9$], and add it to equation $[3]$:
$\quad 3x +3y-9z=-9$
$\underline{+ 2x -3y +8z = a}$
$\;\;5x +0y-z=-9 +a \implies 5x -z = -9 +a\qquad$ [II]
So we're left with [I] $5x - z = 4$ and [II] $5x - z = -9 + a$.
For [I] and [II] to both hold, we must have that $4 = -9 + a$, from which we can solve for $a$ by adding 9 to both sides of the equation, giving us: $a = 13$.
The goal, using this approach, is to eliminate a variable by adding/subtracting equations (as with equation $[1]$ and $[2]$), or adding/subtracting a multiple of an equation (as we did when we multiplied equation $[2]$ by $3$, and then added to equation $[3]$).
This is essentially what row reduction will be like, except you won't have the variables cluttering everything up!
A: Yes there is. It is called the Row Echelon form. Row echelon form is sometimes called reduced row echelon form or Hermite normal form.  A matrix is in row echelon form if:


*

*the first nonzero element in every row is 1,

*the first nonzero element in every row occurs to the right of the first nonzero element in the row above it, and

*all the elements above the first nonzero element in a row are 0.


For your case you construct the following from the coefficients and the right hand side:
$$ \begin{bmatrix}\begin{array}{ccc|c}
4 & -1 & 2 & 7\\
1 & 1 & -3 & -3\\
2 & -3 & 8 & a\end{array}\end{bmatrix} $$
and try to eliminate the lower triangular elements. To do so make the first column equal to 4 in all the rows by multiplying by 4 and 2 rows 2 and 3 to get 
$$ \begin{bmatrix}\begin{array}{ccc|c}
4 & -1 & 2 & 7\\
4 & 4 & -12 & -12\\
4 & -6 & 16 & 2a\end{array}\end{bmatrix} $$
and then subtract the 1st row from all the other rows to get 
$$ \begin{bmatrix}\begin{array}{ccc|c}
4 & -1 & 2 & 7\\
0 & 5 & -14 & -19\\
0 & -5 & 14 & 2a-7\end{array}\end{bmatrix} $$
The repeat for the 2nd column starting from the 2nd row and down. This gets us to
$$ \begin{bmatrix}\begin{array}{ccc|c}
4 & -1 & 2 & 7\\
0 & 5 & -14 & -19\\
0 & 0 & 0 & 26-2a\end{array}\end{bmatrix} $$
which obviously cannot be solved for $x$, $y$, $z$ at the same time because all their coefficients are zero in the last equation.
The above is expanded to
$$ \begin{aligned}4\, x-y+2\, z & =7\\
5\, y-14\, z & =-19\end{aligned} $$
which is solved for $x$ and $y$ in terms of $z$ for example, and
$$ 2\,a = 26 $$.
A: Another way of thinking about it: you really have 3 equations in 4 unknowns $x$,$y$,$z$,$a$ (except that you only care about the value of $a$ in the solution).  So do row reduction on the matrix
$\left[ \matrix{4 & -1 & 2 & 0 & 7\cr 1 & 1 & -3 & 0 & -3\cr
2 & -3 & 8 & -1 & 0\cr} \right]$.  
A: If you look at the augmented matrix, and start to row reduce, it only takes two reductions (depending how you count) before you can see that $6+a=19$.
$\begin{bmatrix} 4 & -1 & 2 & 7\\ 1 & 1 & -3 & -3\\ 2 & -3 & 8 & a\end{bmatrix}\rightarrow^{-4\textrm{R}_2+\textrm{R}_1\rightarrow\textrm{R}_1}_{-2\textrm{R}_2+\textrm{R}_3\rightarrow\textrm{R}_3}\rightarrow\begin{bmatrix} 0 & -5 & 14 & 19\\ 1 & 1 & -3 & -3\\ 0 & -5 & 14 & 6+a\end{bmatrix}$
