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Fairly basic algebra question - but it's bugging me now!

  1. Dec 9, 2006 #1


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    I'm just going through some notes on linear interpolation, but this bit has got me confused - I think it's an easy work through on gradients etc, but my maths seems to be failing me (mature student here so go easy on me :shy: )

    Ok here goes (see attached images - not sure about latex yet. A look at image 2 first, it just shows the gradient equation)

    My question is: How can e.g. (x2-x) become (x-x2), I was always taught that when you move the lower part of a division/fraction to the other side of the equals sign, you don't change it! So, how does this work?

    Is there some trick that occurs when you rearrange an equation or is there some maths rule that I was never taught.

    I know this is probably a solve in 2 seconds type question, but it's starting to bug me now !!:grumpy:


    Attached Files:

  2. jcsd
  3. Dec 9, 2006 #2


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    Dearly Missed

    Do not post your questions in the tutorial section!
    A mentor will move this thread to where it belongs.
  4. Dec 9, 2006 #3


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    Whoops! Sorry, forgot which area I was in. o:)

    Please moderator (any one of you) please could you move this to the relevant area.

  5. Dec 9, 2006 #4


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    Well, there is NO (x2-x) that becomes (x-x2)! You mean "How can (x2-x1) become (x1-x2).
    You have
    [tex]y= \frac{y_2-y_1}{x_2-x_1}(x- x_1)+ y_2[/tex]
    and it becomes
    [tex]y= y_1\frac{x-x_1}{x_1-x_2}+ y_2\frac{x-x_1}{x_2-x_1}[/tex]
    Actually that is wrong but not because of the switch of (x2-x1) to (x1-x2).
    What you do have is
    [tex]\frac{y_2-y_1}{x_2-x_1}(x- x_1)= y2\frac{x-x_1}{x_2-x_1}- y_1\frac{x-x_1}{x_2-x_1}[/itex]
    Do you see that it is negative y1 times the fraction? Take that negative into the denominator of the fraction:
    [tex]-y_1\frac{x-x_1}{x_2-x_1}= y_1\frac{x-x_1}{x_1-x_2}[/tex]

    What is wrong is that the last "y2[/sup]" has been dropped! did you mean
    [tex]y- y_2= y_1\frac{x-x_1}{x_1-x_2}+ y_2\frac{x-x_1}{x_2-x_1}[/tex]
    [tex]y- y_2= y_1A_1(x)+ y_2A_2(x)[/tex]?

  6. Dec 10, 2006 #5


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    Actually, I just lifted it straight from the course text (pdf file). So any errors will be in the original teaching notes.

  7. Dec 10, 2006 #6


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    So basically I can take the minus sign down to the denominator, but then I have to switch the numbers in the demominator to keep the value of the overall multiplication the same ie -2(1/3-1) = 2(1/1-3), I worked these both to be -1, so I guess it works.

    Can I also ask, how does this work when you have a long denominator ie -2(1/(2+3-5)*2) can you still move the minus sign down? do I just have to change the +'s to -'s or would I have to rearrange.

    Thanks again
  8. Dec 10, 2006 #7


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    You can "move" a minus sign wherever you want:
    [tex]-\frac{a}{b}= \frac{-a}{b}= \frac{a}{-b}[/tex]
    In the example you give
    [tex]-2\frac{1}{(2+ 3-5)^2}= \frac{2}{-(2+3-5)^2}= \frac{-2}{(2+3-5)^2}[/itex]
    Of course, sinced 2+3-5= 0 that fraction doesn't exist wherever the "-" is! (Were you trying to catch me?)
  9. Dec 10, 2006 #8


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    Nope, just made that question up on the spot.

    Thanks for your help
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