Thread: Actions on a Solution Set

Hi, can anyone help me explain why these actions do not change the solution set of a system of equations?

i. Multiplying one equation by a non-zero number
ii. Adding one equation to another

How do I explain this if S is the solution set of the system of m equations with the unknowns and where if each equation in the system is satisfied, then

Thank you.

I'm a bit puzzled as to why you would be asking. If you are dealing with systems of equations, you must have years of experience with solving individual equations and what you ask is just what you learned long ago:

If you multiply both sides of a true equation by a number, the equation is still true. Any equation is of the form A(x, y, z, ...)= B(x, y, z, ...), where x, y, z, ... are the variables, which is equivalent to A(x, y, z, ...)- B(x, y, z, ...)= 0. For any number, u, u times 0, u(0)= 0 so specifically, u(A(x, y, z, ...)- B(x, y, z, ...))= uA(x, y, z, ...)- uB(x, y, z, ...)= 0. That is the same as u(A(x,y,z,...)= uB(x, y, z, ...)).

Similarly, if A(x, y, z, ...)= B(x, y, z,...) and C(x, y, z, ...)= D(x, y, z, ...), then A(x, y, z, ...)- B(x, y, z, ...)= 0 and C(x, y, z, ...)- D(x, y, z, ...)= 0.

0+ 0= 0 so (A(x, y, z, ...)- B(x, y, z, ...))+ (C(x, y, z, ...)- D(x, y, z, ...))= 0. Then (A(x, y, z, ...)+ C(x, y, z, ...))- (B(x, y, z, ...)+ D(x, y, z, ...))= 0 so A(x, y, z, ...)+ C(x, y, z, ...)= B(x, y, z, ...)+ D(x, y, z, ...).