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Systems of linear equations

  1. Mar 3, 2009 #1
    1. The problem statement, all variables and given/known data
    I wish to solve the following two equations, which are related:

    [tex]
    h(x,y) = y \quad \text{and} \quad g(x,y) = y+x^2-bx.
    [/tex]

    First I set h=0, which gives me y=0. Then I insert y=0 in g, where I find x=b and x=0. All is good here.

    *********

    Now I wish to solve the same system, but I write it as follows:

    [tex]
    h(x,y) = y =0\cdot x + y \quad \text{and} \quad g(x,y) = y+x^2-bx.
    [/tex]

    Here I see that all x and y=0 satisfy h(x,y)=0, why I insert these values in g. But all x and y=0 do not satisfy g(x,y)=0 now. I know my reasoning is wrong somewhere, but I cannot see where. Can you shed some light on this simple problem?

    Thanks in advance.
     
  2. jcsd
  3. Mar 3, 2009 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    I don't understand what you asking or what "problem" you have but a few comments. First, you titled this "Systems of linear equations" but the second is not a linear equation. Second, you set "h= 0", why? Are you saying that you actually are solving "y= 0, y+ x2- bx= 0"? If so why define "h" and "g" at all?

    Finally, I don't understand why you think that "all x and y= 0" should satisfy both equations. Adding a second equation restricts the possible solutions. Any solution to the both equations must satisfy the first equation by not the other way around: some solutions to one equation will satisfy both. Yes, "all x and y=0" satisfy y= 0. That's true whether you write "+ 0*x" or not. Geometrically, that is the vertical, y, axis. y+ x2- bx= 0 is the same as y= bx- x2 and is satisfied by all points on the parabola. Points that satisffy both are the points where they intersect: x= b, y= 0 and x= 0, y= 0.
     
  4. Mar 3, 2009 #3
    My question is that when I have to solve a non-linear and/or linear system of equations, then what is the most efficient way of doing it, and when I am I allowed to insert the solution that satisfies 1 of the equations into the others.

    1) Yeah, the topic should have been non-linear. My mistake.

    2) Yes, I guess it is overkill defining h(x,y) and g(x,y) in the first place.

    3) If adding a second equation restricts the possible solutions, then how is it that in the first example in post #1 I can just insert y=0 in the second equation (i.e. the equation for g(x,y))?

    Thanks for helping.
     
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