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Math Help - Too many equations. Can I eliminate one this way?

  1. #1
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    Too many equations. Can I eliminate one this way?

    Hello all,

    I have a big long system of equations I am trying to solve and I have more equations than unknowns... let me tell you how I got there. Lets say I am trying to solve for x, y, w/x and z/y, and right now I have 4 equations (we are fine right now, but x and y don't mean much in the context I am working in). Two of these four equations are...

    y = a x + b z
    x = c y + d w

    I divide both sides of the top equation by y and both sides of the bottom equation by x. Now I have

    1 = a (x/y) + b (z/y)
    1 = c (y/x) + d (w/x)

    Okay so now I am just trying to solve for x/y w/x and z/y (but I have four equations). There is no reason why I cant just set a (x/y) + b (z/y) equal to c (y/x) + d (w/x) to eliminate this extra equation right?

    Cheers!

    Nick
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  2. #2
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    Quote Originally Posted by salohcin View Post
    Hello all,

    I have a big long system of equations I am trying to solve and I have more equations than unknowns... let me tell you how I got there. Lets say I am trying to solve for x, y, w/x and z/y, and right now I have 4 equations (we are fine right now, but x and y don't mean much in the context I am working in). Two of these four equations are...

    y = a x + b z
    x = c y + d w

    I divide both sides of the top equation by y and both sides of the bottom equation by x. Now I have

    1 = a (x/y) + b (z/y)
    1 = c (y/x) + d (w/x)

    Okay so now I am just trying to solve for x/y w/x and z/y (but I have four equations). There is no reason why I cant just set a (x/y) + b (z/y) equal to c (y/x) + d (w/x) to eliminate this extra equation right?

    Cheers!

    Nick

    Certainly you have 4 equations but two of them are directly derived from the other two, and you still have 3 unknowns:

    putting \displaystyle{A:=\frac{x}{y}\,,\,\,B:=\frac{z}{y}\  ,,\,\,C:=\frac{w}{x}}, the last two equations you have are

    \displaystyle{aA+bB=1\,,\,\,\frac{c}{A}+dC=1} ....you're missing one (independent) equation if you want to solve this.

    Tonio
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