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Graph algebra

  1. Aug 1, 2014 #1
    In need of help. Ill write it is how it is in the text book.

    Find the equation of the line having gradient 3/4, that passes through 7,11.
    Express your answer in the form i) ax + by + c =0 and ii) y = mx + c

    As one point and the gradient are known, use the formula: y - y1 = m(x-x1)

    Plug in numbers is y-11 = 3/4(x-7)
    Simplify expressing in the form ax + by + c = 0

    y-11 = 3/4(x-7)
    4y-44 = 3(x-7)
    4y-44=3x -21
    Heres where im really baffled next step is
    3x-4y+23 = 0

    In algebra dont you do what you do to the right what you do to the left???? How does moving a positive 3x in front of the 4y put a minus sign in front of 4y. What am i doing here exactly??
     
  2. jcsd
  3. Aug 1, 2014 #2

    HallsofIvy

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    I've always disliked the idea of "moving" something from one side of an equation to the other! "Moving" a number or expression is NOT an arithmetic or algebraic step.

    What you are doing is subtracting 4y from both sides of the equation and adding 21 to both sides of the equation. Do those two things to 4y- 44= 3x- 21 and see what you get.
     
  4. Aug 7, 2014 #3

    symbolipoint

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    One can say, "apply the appropriate additive and multiplicative inverses to have the variable on one side and just the constants on the other side". Nice but very wordy.
     
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