Thread: Postulate for elimination?

In proving that the elimination method of adding linear equations works, does one need to show that it is equivalent to substitution?
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?

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? 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.

Originally Posted by lamp23

In proving that the elimination method of adding linear equations works, does one need to show that it is equivalent to substitution?

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.

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?

Other than that ERO's don't change the solution space, you mean?

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?

I suppose so. It doesn't seem like a good idea to introduce unnecessarily complicated notation, though.

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.

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.

Maybe my problem is similar to knowing exactly how we get from m = (y_2 - y_1) / (x_2 - x_1) to:
y - y_1 = m(x - x_1)

(We just multiply through by the denominator and change our second point from fixed to variable.) I bring this up because such a subtle notation can be a big idea.


When we set up systems of equations as:

4x + 5y +8z = 2
5x + 4y +9z = 6

are we not implicitly setting the y's equal to each other? Are problems that label equations as y=x^2 and then y=x+4 on the same graph being a little sloppy? It seems like maybe we are not using distinct notation precisely so that an explicit step of setting the y's equal to each other is unnecessary.

P.S. I think MIT's OCW Calculus Lecture 1 actually answers my question:
Here is where Jerison talks about the "sloppiness" (even if it is called single variable calculus because only one variable is being differentiated at a time (right?)).

Originally Posted by lamp23

Maybe my problem is similar to knowing exactly how we get from m = (y_2 - y_1) / (x_2 - x_1) to:
y - y_1 = m(x - x_1)

(We just multiply through by the denominator and change our second point from fixed to variable.) I bring this up because such a subtle notation can be a big idea.


When we set up systems of equations as:

4x + 5y +8z = 2
5x + 4y +9z = 6

are we not implicitly setting the y's equal to each other?

Nothing implicit about it. Usually, such systems are called "simultaneous systems of equations", for the full name. The word "simultaneous" there means that the x's, the y's, and the z's are precisely the same variable, representing precisely the same thing.

Are problems that label equations as y=x^2 and then y=x+4 on the same graph being a little sloppy?

Technically, I suppose you're right. A more precise method would be to say that and Your variable typically would be the same, but you mean different functions of that variable

You have to depend on context for the information. Mathematical language, if done well, is less ambiguous than English (or any other spoken language, for that matter), but it still can have some ambiguities - ambiguities that resolve themselves in context.

It seems like maybe we are not using distinct notation precisely so that an explicit step of setting the y's equal to each other is unnecessary.

I suppose so. Authors can be lazy sometimes, and want their readers to do more work so they don't have to do quite as much. That's not necessarily a bad thing. Readers always have to do some work (Just read How to Read a Book, by Mortimer Adler, to see just how much work a good reader must do!), which is a good thing. You'd hardly learn from books if you didn't have to work at it some.

Originally Posted by Ackbeet

Nothing implicit about it. Usually, such systems are called "simultaneous systems of equations", for the full name. The word "simultaneous" there means that the x's, the y's, and the z's are precisely the same variable, representing precisely the same thing.

If we are trying to find the point of intersection of two lines, and , then isn't this a case where by writing exactly and exactly in each different line that we are in effect implicitly setting the two different y variables equal and the two different x variables equal--the purpose of which is for a simplified technique when using either substitution or elimination?

Originally Posted by lamp23

If we are trying to find the point of intersection of two lines, and , then isn't this a case where by writing exactly and exactly in each different line that we are in effect implicitly setting the two different y variables equal and the two different x variables equal--the purpose of which is for a simplified technique when using either substitution or elimination?

Come to think of it, I suppose there are times when x and y are not identically the same thing, a priori. If you're trying to find the intersection of two lines, then the x's would be identical, a priori, but the y's would not. On the other hand, if you're trying to solve a system of equations for x and y, then typically x and y represent precisely the same entities in both equations.

Practically speaking, of course, these subtleties usually make little difference in the mechanics of solution, although I suppose for full understanding of the problem statement, it wouldn't hurt to spell it out.