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Ax+b=0 is one-variable linear equation

  1. Mar 13, 2014 #1
    Am I right in saying:
    ax+b=0 is one-variable linear equation
    ax+by+c=0 is two-variable linear equation
    ax^2+bx+c=0 is one-variable quadratic equation
    ax^2+bx+c=y is two-variable quadratic equation
    Every linear or quadratic equation in one or two variables can be represented in those ways.

    How come graph of ax+by+c=0 is point of solution set {(x,y) | ax+by+c=0}? Why not (y,x)? Since we can rearrange it as x=(y=b)/m, can we pick the y value first and then find the x value?

    I know graph of x=a and y=a is vertical and horizontal lines respectively because x or y will always be constant no matter what y or x is respectively. But how would we represent the solution sets? For x=a, would it be {(x,y) | x+0y=a}?

  2. jcsd
  3. Mar 13, 2014 #2

    I would say the general two-variable quadratic equation has the form

    [tex]ax^2 + by^2 + cxy + dx + ey + f = 0[/tex]

    It doesn't matter much. You might as well choose

    [tex]\{(y,x)~\vert~ax+ by + c = 0\}[/tex]

    It might not be the same set, but it will be an equivalent situation you're dealing with. Geometrically, this corresponds to just swapping X-axis and Y-axis.

    Usually, we will of course use

    [tex]\{(x,y)~\vert~ax + by + c = 0\}[/tex]

    but that's a convention.

    Yes, that will be one possible way of writing the solution set. Another one is

    [tex]\{(x,y)\in \mathbb{R}^2~\vert~ x =a\}[/tex]


    [tex]\{(a,y)\in \mathbb{R}^2~\vert~y\in \mathbb{R}^2\}[/tex]
  4. Mar 13, 2014 #3
    Thanks a lot. That cleared my doubts.
  5. Mar 14, 2014 #4
    Am I right in saying:
    graph of f(x) is graph of solution set {(x,f(x)) | f(x)=x+a}

    Can I right it as x=f(x)-a too?
  6. Mar 14, 2014 #5
    I would say the graph of a function ##f:\mathbb{R}\rightarrow\mathbb{R}## is

    [tex]\{(x,f(x))\in \mathbb{R}^2~\vert~x\in \mathbb{R}\}[/tex]

    or perhaps

    [tex]\{(x,y)\in \mathbb{R}^2~\vert~y=f(x)\}[/tex]

    In the case that ##f(x) = x+a## for all ##x##, then this becomes

    [tex]\{(x,x+a)\in \mathbb{R}^2~\vert~x\in \mathbb{R}\}[/tex]


    [tex]\{(x,y)\in \mathbb{R}^2~\vert~y=x+a\}[/tex]

    and you can write this of course as

    [tex]\{(x,y)\in \mathbb{R}^2~\vert~x=y-a\}[/tex]
  7. Mar 14, 2014 #6
    [tex] {(x,f(x)) | f(x)=x+a} [/tex]
    is wrong then?

    Can't get { and } to show...
  8. Mar 14, 2014 #7
    If your ##f## is defined by ##f(x) = x+a##, then that just says


    which simplifies to


    So yes, it's not really right.
  9. Mar 14, 2014 #8
    The how come

    [tex] (x,y) | y=x+a [\tex]

    is correct?
  10. Mar 14, 2014 #9
    Then how come

    [tex] (x,y) | y=x+a [/tex]

    is correct?
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