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Math Help - Simultaneous Equation [C1]

  1. #1
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    Simultaneous Equation [C1]

    I've been going through a textbook and I've came across Simultaneous equations by this point. I can do many, such as:

    c = 7d - 2 [1]
    3c - 4d = 11 [2]

    To do this I subbed c in [2] and did this method:

    3(7d-2) - 4d = 11
    21d - 4d - 6 = 11
    17d = 17
    d = 1

    And then went back to 3c - 4d = 11
    I found that 3c = 15
    and c = 5.

    However the last question is:

    6x - 5y = 27
    3y = 1 - 5x

    Now I'm pretty sure you don't use the same method as last time so what can I do here
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  2. #2
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    For example: from the second equation: y=\frac{1-5x}3 and then substitute this into the first one: 6x-\frac 53(1-5x)=27\leftrightarrow 18x-5(1-5x)=81\leftrightarrow 18x-5+25x=81\leftrightarrow 43x=86\leftrightarrow x=2 substitute this into y=\frac{1-5x}3:  y=\frac {1-5\cdot 2}3=-3.
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  3. #3
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    rearrange so that the x and y on same side

    6x - 5y = 27
    3y + 5x = 1

    then multiply top equation by 3(for y) and bottom equation by 5 so you get

    18x - 15y = 81
    15y + 15x = 5

    now i think you will be able to solve it
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  4. #4
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    i made a slight mistake, the bottom equation should say 15y + 25x = 5
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