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Math Help - Solving simultaneous eqautions by inspection

  1. #1
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    Solving simultaneous eqautions by inspection

    Hi,

    I have attached what I am trying to do hopefully it is clear any ideas?

    Thanks
    Attached Thumbnails Attached Thumbnails Solving simultaneous eqautions by inspection-picture.png  
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  2. #2
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    Subtract the 2nd equation from the first equation and you have

    -2y = -2 so y = 1 then substitute y = 1 into 1st equation and you have

    x + 8(1) = 4 so x = -4

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  3. #3
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    Quote Originally Posted by Caity View Post
    -2y = -2 so y = 1
    Thanks for that but why does -2 become y=1, I would think logically it was y=-2
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  4. #4
    A riddle wrapped in an enigma
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    x+8y=4

    x+10y=6

    Subtract the two equations.

    -2y=-2

    Divide both sides of the equation by -2.

    \frac{-2y}{-2}=\frac{-2}{-2}

    y=1

    Substitute y = 1 into the first equation and solve for x.
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  5. #5
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    Thanks that really helped but I got stuck again on this one

    5x-3y=47 1

    7x-3y=67 2

    So we ignore the 3y as (-3y) - (-3y) is 0.

    On the right: 47-67 = -20
    On the left: 5x-7x = -2x

    Therfore:
    \frac{-2x}{-20} = 10x

    Now we substitute x = 10 into formulae 1, because it has the lowest number on right

    5 x 10 = 47

    50 = 47

    47 - 50

    = -3

    y = -3

    Although the answer says y = -1

    Hope this is clear, and thank you.
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  6. #6
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    You have it correct up to -3y = -3
    then you divide both sides by the -3 because you want y by itself.
    so y= 1 not -1. sometimes the answers in the book have typos
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  7. #7
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    Hey thanks for that. Ive just got to the bit where the y or x's are more than 1 never knew u had to devide ... Thanks
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