Does exist a proof that the well known equivalent manipulation in linear equations are really always equivalent?

Question

Linear equations can be written as two functions: $f(x) = g(x) $. Does exist a proof that for example if we add a same number to both sides of a linear equation, then the functions on both sides of the equations will be still equivalent? Does exist a proof for other equivalent manipulations?

If I write for example an equation $$x+2=1/2$$, then these two functions are very different from functions $$2x+4=1$$, yet the intersection of those two functions has the same coordinate X as those two functions before. What can give me confidence that I can do this operation for both sides of the equations and really always has the same coordinate X for the intersection of those two functions no matter what number I multiply it?

Answer 1

There is two answer I can give:

Historical You are right! It's really not clear that we can do that. We do it because they teach us, we not comprehend it! In fact in past ages of mathematics history nobody do it! Khwarizmi is the well known iranian mathematician that invented it and in fact the Algebra. It wrote the book "The Compendious Book on Calculation by Completion and Balancing" that is the start of modern algebra. Before that also no equation wrote by unknown variables! Just describe the problem and try to solve it by words not equations. For more details you can search for Khwarizmi and his book.

Algebraic Also may be look the equations in a additive group, and then use cancelation law in groups. But I don't know it satisfy you or NOT!

For myself the historical review is more satisfactory. I very believe that math history must be one of the basic and mandatory lessons that every term any math student should has one. Continue the last steps of our teachers without any sight of how we get to this point and how human being thought was, is very senseless.

Answer 2

Addition is a function, hence if $a=b$, then $a+c=b+c$, and conversely, by adding $-c$.

Hence, any solution to $f(x)=g(x)$ is also a solution to $f(x)+c=g(x)+c$ and conversely, and a solution to $f(x)\ne g(x)$ is also a solution to $f(x)+c\ne g(x)+c$ and conversely.