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Math Help - Actions on a Solution Set

  1. #1
    Junior Member
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    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 x_{1}, x_{2}, ... , x_{n} and \overline{s} = s_{1}, s_{2}, ... , s_{n} \in S where if each equation in the system is satisfied, then x_{1} = s_{1} , ... , x_{n} = s_{n}

    Thank you.
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  2. #2
    MHF Contributor

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    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, ...).
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