Adding equations in the elimination method

[Vay]

So my textbook says multiply one of the linear equation until it can cancel out with another linear equation to get the solution for a variable, when the two equations are added. My question is, what is the proof or justification for adding equations that arrives to a solution? In other words, is this a mere coincidence that by adding the two equations, we can get the answer for a variable?

Edited November 8, 2011 by Vay

[DrRocket]

So my textbook says multiply one of the linear equation until it can cancel out with another linear equation to get the solution for a variable, when the two equations are added. My question is, what is the proof or justification for adding equations that arrives to a solution? In other words, is this a mere coincidence that by adding the two equations, we can get the answer for a variable?

 

 

 

If a=b and c=d then a-c=b-d. QED

[timo]

As for a "proof" (more an explanation in fact): Both sides of the equation are the same (that is the definition of an equation). So you add the same value to both sides of equation (1), merely represented in different forms (namely the left- and the right-hand side of equation 2). So on the left-hand side of equation (1) you have a value and add some other value, and on the right-hand side you add the same two values. After addition, both sides of course are equal again. Note that the same reasoning applies for multiplying an equation with a constant: Both sides are equal before multiplication, so they are after both have been multiplied with the same constant. (*)

EDIT: Above is really just the very elaborated form of what DrRocket said.

 

It is not coincidence that by adding the equations you successively eliminate variables. You explicitly chose your factors such that that happens. It is also not a complete coincidence that this method works: You are not doing this on arbitrary equations but (usually) only on linear equations of the type ax+by+... = d (i.e. no equations like tan(x) + 5y² = d).

 

 

(*) Careful: Two sides being equal after a multiplication does not always imply that they were equal before the multiplication, since [math] 1\neq 2 [/math] but [math] 0 \cdot 1 = 0 \cdot 2 [/math].

Edited November 8, 2011 by timo

[Vay]

I just figured out that we assume that a system of linear equations has a solution, then we proceed to solve it. But depending on the answer, then can we really claim whether it has a solution.

 

Also the fact that a=b c=d

 

then a+b=b+d as well, thanks.

Edited November 8, 2011 by Vay