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An inquiry on solving systematic equation

  1. Nov 20, 2011 #1
    Let's say we have two equations:
    ax+by+c=0
    dx+ey+f=0
    We often find the point of intersection by substitution or elemination.
    If we were to solve by elemination, we either add or substract the two equations (or manipulate them) to eleminate one variable, in order to express the other variable only in terms of non-variable (that is, fixed point).

    What does it mean? How does a set of element that satisfies a equation be manipulated to find a subset of the set of which the subset also satisfies the other equation?

    Or what does adding to equation mean 'graphically'? (I am certain that a better word exists)
    I somehow can feel that adding or subtracting a non-variable to a equation, but I don't understand the significance of adding or subtracting equations
     
  2. jcsd
  3. Nov 21, 2011 #2

    Simon Bridge

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    Math is language and algebra is grammar - the equations are sentences: you've heard them called "expressions".

    When you manipulate an equation, you are rewriting the sentence. Just like with sentences, this can change the meaning of the sentence. The rules of algebra tell us how to manipulate a sentence without changing it's meaning - we make a tautology.

    The point of doing this is to make relationships between different parts of the expression clear. You'll have noticed that the form: y=mx+c is easier to graph than, say, ay+bx+k=0 ... but these are just two ways of describing the same kind of object - which we can prove by following the rules for algebra. In fact, we can show that both these expressions will describe exactly the same line provided some conditions are met relating the coefficients.

    The simultanious equations problem basically starts out by describing two sets of points - one for each equation. You are asked to find what points these sets have in common.

    The set of points they have in common is called the intersection.

    Where the equations are straight lines, as in your example, the intersection will either be empty (they are parallel), all the points (they are the same line), or just one point.

    Does that help?
     
  4. Nov 21, 2011 #3
    Seems like I didn't cleary express my self.

    'adding a non-variable to an expression' means a 'transition' to certain direction. Say y=2x+3. If we add 5 to the right side of equation, we are moving the line 5 units upwards.

    But I don't understand the rationale of adding variable to the equation to find a intersection point, or solving a systematic equation by substitution or elimination.
     
  5. Nov 21, 2011 #4
    Do you have any experience in linear algebra? Things like elimination (row operations) have very nice explanations in terms of subspaces and linear transformations.
     
  6. Nov 21, 2011 #5
    No, I am a high school student
     
  7. Nov 21, 2011 #6
    If you're interested in "reading ahead", you may try checking out Gilbert Strang's lectures on MIT OpenCourseWare (you can find them http://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2010/" [Broken]). Essentially, MIT has put a number of courses online, video lectures included. Strang's linear algebra course will give you a very good idea of what's going on when you deal with systems of equations.
     
    Last edited by a moderator: May 5, 2017
  8. Nov 21, 2011 #7

    AlephZero

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    It means you are creating a new equation that has the same solution(s) as the old ones.

    In other words, if are thinking of the two equations as lines on a plane and the solution as the point of intersection, a linear combination of the equations is another line that goes through the same intersection point.

    If the solution of your equations is (x0, y0), then
    c = -ax0 -by0
    f = -dx0 - ey0
    and so
    a(x-x0) + b(y-y0) = 0
    d(x-x0) + e(y-y0) = 0

    and now you can see that any linear combination of the equations is also of the form
    P(x-x0) + Q(y-y0) = 0
    for some constants P and Q.

    In other words, it is the equation of a line through (x0,y0) with slope -Q/P.
     
  9. Nov 21, 2011 #8
    thanks!
     
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  10. Nov 21, 2011 #9

    Simon Bridge

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    The others have given you some algebra courses to try.
    I noticed that the way you wrote suggests you think of math operations as movements in space (or that this is a way that helps you?) I thought I'd try a description that used this:

    First, I'll recap what you said, make sure I understand you,
    then I will add a variable to an equation and show you what this does to it,
    then I'll expand this idea to cover adding whole expressions together in different combinations.

    Set:
    y=2x+3

    Adding a constant:
    If we transform: y'=y+5
    then y'=2x+8
    ...which is the original line translated 5 units in the y direction.

    Adding a variable:
    If we transform: y'=y+x
    then y'=3x+5
    ...which is the original line rotated clockwise a bit.

    We can also add two equations to each other:

    (y=2x+3)+(y=4x+5)-->(2y=6x+8)-->(y=3x+4)

    This rotates the lines about the point that they intersect.
    (sketch all three and you'll see it.)

    This particular result is not terribly helpful when it comes to finding the intersection.
    In general you can add multiples like this:
    a(y=4x+5) + b(y=2x+3)
    ... to produce an arbitrary rotation about the intersection.

    To find the intersection, what we want to do is rotate the line about the intersection until it is horizontal, so the equation is y=1 (in this case) and so you know the y-position of the intersection. From there finding the line x=-1 (in this case) is trivial.

    Sounds rough but there is a trick to it. You learn the trick.
    What you are probably being taught at High School is a collection of tricks for different classes of simultaneous equations. For the example, if you put a=1 and b=-2 this happens:

    1(y=4x+5) + -2(y=2x+3) = (-y=0x-1) so (y=1) see?

    (again, a sketch will show you what is happening)
    This is "elimination" of x: it rotates the line about the intersection so it is horizontal (eliminate y and you get a vertical line.)

    Of course, you could do y'=y-4x in the first equation - that will give you a horizontal line too. Unfortunately it did not rotate about the intersection so y=5 (means x=0) will not be the correct intersection.


    Thinking of these operations as movements in space is nice going, but will trip you up when you deal with more abstract math. Best practice is to get used to different ways of thinking about math.

    How someone would normally do the last example, is to realize that y has to have the same value in both equations and reason that the only way that happens is if
    (4x+5)=(2x+3)

    ... (just putting y=y) so you write that down then use the rules of algebra to make x the subject (called "solving for x"). This way looks for the vertical line first. (Eliminating y, by putting a=1 and b=-1 in the first method.)

    It is equally valid to make x the subject of both equations, and put x=x:
    (y-5)/4 = (y-3)/2

    Also equally valid to make x the subject of one, then substitute it into the other, then solve for y. Which sounds more complicated but is usually easier to do:
    x=(y-5)/4 ...(from the 1st equation, sub into the second)
    y=2(y-5)/4 +3

    These are all doing the same things.
    Sometimes it is not immediately obvious which a and b to use when adding equations (first method) together - so we break the process down into easier steps ... which is what the last three methods were doing. We pick the method that makes the math easier for us.
     
    Last edited: Nov 21, 2011
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