TITLE: Systems of linear equations: Why does no one plug back in?
QUESTION [49 upvotes]: When someone wants to solve a system of linear equations like
$$\begin{cases} 2x+y=0 \\ 3x+y=4 \end{cases}\,,$$
they might use this logic: 
$$\begin{align}
\begin{cases} 2x+y=0 \\ 3x+y=4 \end{cases} 
\iff &\begin{cases} -2x-y=0 \\ 3x+y=4 \end{cases} 
\\ 
\color{maroon}{\implies} &\begin{cases} -2x-y=0\\ x=4 \end{cases} 
\iff \begin{cases} -2(4)-y=0\\ x=4 \end{cases} 
\iff \begin{cases} y=-8\\ x=4 \end{cases}
\,.\end{align}$$
Then they conclude that $(x, y) = (4, -8)$ is a solution to the system. This turns out to be correct, but the logic seems flawed to me. As I see it, all this proves is that 
$$
  \forall{x,y\in\mathbb{R}}\quad
  \bigg(
  \begin{cases} 2x+y=0 \\ 3x+y=4 \end{cases} 
  \color{maroon}{\implies}
  \begin{cases} y=-8\\ x=4 \end{cases}
  \bigg)\,.
$$
But this statement leaves the possibility open that there is no pair $(x, y)$ in $\mathbb{R}^2$ that satisfies the system of equations.
$$ 
\text{What if}\; 
\begin{cases} 2x+y=0 \\ 3x+y=4 \end{cases} 
\;\text{has no solution?}
$$
It seems to me that to really be sure we've solved the equation, we have to plug back in for $x$ and $y$. I'm not talking about checking our work for simple mistakes. This seems like a matter of logical necessity. But of course, most people don't bother to plug back in, and it never seems to backfire on them. So why does no one plug back in?
P.S. It would be great if I could understand this for systems of two variables, but I would be deeply thrilled to understand it for systems of $n$ variables. I'm starting to use Gaussian elimination on big systems in my linear algebra class, where intuition is weaker and calculations are more complex, and still no one feels the need to plug back in.

REPLY [7 votes]: Another point of view.  Gaussian elimination of a system of linear equations is equivalent to multiplication by certain types of elementary matrices.  To avoid getting bogged down I will give specific examples and get you to look up more general cases if you find it of interest.
There are three basic types of row operation.

Add a multiple of an equation to another equation, say, $3$ times equation 1 to equation 2.  For example
  $$\eqalign{x+2y&=3\cr4x-5y&=6\cr}\quad\longrightarrow\quad
  \eqalign{x+2y&=3\cr7x+\phantom{1}y&=15\cr}$$
  This corresponds to multiplying the (augmented) coefficient by a certain matrix,
  $$\pmatrix{1&2&3\cr4&-5&6\cr}\quad\longrightarrow\quad
  \pmatrix{1&0\cr3&1\cr}\pmatrix{1&2&3\cr4&-5&6\cr}
  =\pmatrix{1&2&3\cr7&1&15\cr}$$
  
  Multiply an equation by a non-zero constant, say, the first row by $-2$.  For example,
  $$\eqalign{x+2y&=3\cr4x-5y&=6\cr}\quad\longrightarrow\quad
  \eqalign{-2x-4y&=-6\cr4x-5y&=6\cr}$$
  corresponds to
  $$\pmatrix{1&2&3\cr4&-5&6\cr}\quad\longrightarrow\quad
  \pmatrix{-2&0\cr0&1\cr}\pmatrix{1&2&3\cr4&-5&6\cr}
  =\pmatrix{-2&-4&-6\cr4&-5&6\cr}$$
  
  Finally, interchange two equations:
  $$\eqalign{x+2y&=3\cr4x-5y&=6\cr}\quad\longrightarrow\quad
  \eqalign{4x-5y&=6\cr x+2y&=3\cr}$$
  which corresponds to
  $$\pmatrix{1&2&3\cr4&-5&6\cr}\quad\longrightarrow\quad
  \pmatrix{0&1\cr1&0\cr}\pmatrix{1&2&3\cr4&-5&6\cr}
  =\pmatrix{4&-5&6\cr1&2&3\cr}\ .$$

And now the point: all three classes of multiplying matrices are invertible, and their inverses are matrices of the same types.  This means you can automatically get from your final equations back to the original, and checking solutions is not necessary.
However, while this works for linear equations, it does not in general work for other types of equations.  Your practice of checking solutions is absolutely correct and very important - please don't stop doing it!!