No, because it's not equivalent. What you want to do is show that each of the three different elementary row operations (ERO's) used in elimination leaves the solution space unchanged. This is, apparently, equivalent to showing that they leave the kernel, or null space, unchanged. Most standard books on linear algebra will include a proof of that fact.

Other than that ERO's don't change the solution space, you mean?Using the transitive or substitution property to solve a system of equations is on solid ground because those can be taken as postulates. However there is no elimination postulate, is there?

I suppose so. It doesn't seem like a good idea to introduce unnecessarily complicated notation, though.Also if you don't write y=x+4 and y=2x like so but instead as

f(x)=x+4 and g(x)=2x

then if you add and get

f(x)+g(x)=3x+4

and then write it as

2f(x)=3x+4

then we are implicitly setting f(x)=g(x) right?

The way I do things, is that I write out the matrix each time, and I just have a little note saying something like which means that I take times row one, add it to row two, and store the result in row two.I guess by using the same letter, y, for each line is just a shortcut for not having to explain what is really going on. Prof. Jerison mentions this "sloppiness" in the first lecture of MIT's OCW Calculus.