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Why are we able to combine terms in an equation?

  1. Jan 3, 2010 #1
    Consider this equation:

    [tex]x^2 + 5 + 4x + 6 + 3x^2 + 3x = 0[/tex]

    We know that we can combine the like terms and write the above equation like this:

    [tex]4x^2 + 7x + 11 = 0[/tex]

    Why were we able to combine the terms?
  2. jcsd
  3. Jan 3, 2010 #2


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    Staff: Mentor

    Why not? (Don't mean to sound flip, but it sure looks obvious.)
  4. Jan 3, 2010 #3
    What you are basically asking are why for numbers a,b and a non-negative integer n we have:
    [tex](a+b)x^n =ax^n + bx^n[/tex]
    In short: because they satisfy this distributive property. If we didn't know this we can deduce it from the distributive property of numbers:
    (a+b)c = ac+bc for all numbers a,b,c.

    For fixed numbers a,b and a non-negative integer n, define the functions f and g by:
    [tex]f(x) = ax^n + bx^n[/tex]
    [tex]g(x) = (a+b)x^n[/tex]
    We wish to show that these are equal, and they are equal if and only if they agree at all function values. So if f(c) = g(c) for all numbers c, but this follows from:
    [tex]f(c) = ac^n+bc^n = (a+b)c^n = g(c)[/tex]
    To show that numbers in general satisfy the distributive property you would have to work from the definition of numbers. For integers this can be worked out from the Peano axioms and an inductive argument. For rationals it follows without much trouble from the property for integers. For reals it depends on the construction, but it can get a bit technical and usually such an argument is shown in either introductory analysis or a rigorous calculus course. For complex numbers it follows from the real case without much trouble.

    One way of summarizing many of these kinds of properties of numbers is to say that it is a field (see http://en.wikipedia.org/wiki/Field_(mathematics)#Definition_and_illustration" for a short list of the properties that make us call a certain set a field).
    Last edited by a moderator: Apr 24, 2017
  5. Jan 3, 2010 #4
    Actually, the question I wanted to ask was this:

    Suppose we have

    [tex]Ax^2 + Bx + C = 4x^2 + 7x + 11[/tex]

    Why do we say here that A = 4, B = 7, and C = 11?
  6. Jan 3, 2010 #5


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    Staff: Mentor

    Because x is an independent variable there. Does that make sense?
  7. Jan 3, 2010 #6
    No, somehow I still don't understand.

    Please don't get angry with me--I have difficulty understanding math concepts.
  8. Jan 3, 2010 #7


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    Staff: Mentor

    No worries! I'm not a math expert, so perhaps there are better words to use than I'm using. Do you see intuitively why it is true, and are looking for a good mathematical proof? The Distributive property comes close, but I suspect that you are looking for something more...
  9. Jan 3, 2010 #8
    I'm not sure what you are asking. If A=4, B =7 and C=11 then certainly:
    [tex]Ax^2 + Bx + C = 4x^2 + 7x + 11[/tex]
    in the same sense that if a=2, then a+1=2+1. Are you asking how we know that these are the unique valid values? In that case subtract the terms on the right hand side and use distributivity to get the equivalent:
    [tex](A-4)x^2 + (B-7)x + (C-11) = 0[/tex]
    and we know that the only real polynomial that is identically 0 is the zero polynomial which has coefficient 0 for all terms so A-4 = 0, B-7 = 0, C-11=0 and therefore:
    A=4, B=7, C=11
  10. Jan 3, 2010 #9
    Because anything around the[tex] Ax^2 + Bx + C[/tex] must be 0. For example, [tex] Ax^2 + Bx = 4x^2 + 7x[/tex]. A has to be 4 because nothing else will be able to get up to 4x^2. The same applies for the other coefficients. That's the logical way if you don't understand the properties.
  11. Jan 4, 2010 #10
    Thank you everyone for explanations. Yes, I was looking for an intuitive and logical explanation rather than a proofy one. I think I now understand somewhat.
  12. Jan 4, 2010 #11
    Remember that "3x" is short-hand for "x + x + x" and that "x^2" is just a shorter way of writing "x * x"

    So when you write x^2 + 5 + 4x + 6 + 3x^2 + 3x

    You are "really" writing the following:

    x*x + 5 + x + x + x + x + 6 + x*x + x*x + x*x + x + x + x

    Since the order you add things in is of no importance (the commutative property of addition) you can sort it as such:

    x*x + x*x + x*x + x*x + x + x + x + x + x + x + x + 5 + 6

    and then when you write that in the shorter form you get

    4x^2 + 7x + 11

  13. Jan 4, 2010 #12
    Good explanation, knewbie! Thanks!
  14. Jan 4, 2010 #13


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    Science Advisor

    Technically, it is the "distributive property": ac+ bc= (a+ c) together with the "commutative property of addition", a+ b= b+ a.

    Buy the commutative property, we can switch those values around to get
    [tex]x^2+ 3x^2+ 4x+ 3x+ 5+ 6= 0[/tex]
    Then, by the distributive property,
    [tex]x^2+ 3x^2= (1+ 3)x^2= 4x^2[/tex]
    [tex]4x+ 3x= (4+ 3)x= 7x[/itex]
    and, of course, 5+ 6= 11.

    Putting those together,
    [tex]x^2+ 4x+ 5+ 3x^2+ 3x+ 6= 4x^2+ 7x+ 11= 0[/tex]
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