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Math Help - Solve each of the systems by substitution.

  1. #1
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    Solve each of the systems by substitution.

    1. 5x 2y = -5 I think I did this wrong can someone help please?
    y 5x = 3


    y = 5x + 3 y =5(-2.2) + 3
    5x 2(5x + 3) = -5 y = -11 + 3 = -8
    5x 10x + 6 = -5
    -5x + 6 = -5
    - 6 = -6
    -5x = -11

    x =-11/-5 = -2.2

    so the solution would be (-2.2, -8)





    2. 4x 12y = 5
    -x + 3y = -1
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Patience View Post
    1. 5x 2y = -5 I think I did this wrong can someone help please?
    y 5x = 3


    y = 5x + 3 y =5(-2.2) + 3
    5x 2(5x + 3) = -5 y = -11 + 3 = -8
    5x 10x + 6 = -5
    -5x + 6 = -5
    - 6 = -6
    -5x = -11

    x =-11/-5 = -2.2

    so the solution would be (-2.2, -8)
    where did you get the -2.2 from, how did it suddenly appear, was there more info in the question?

    5x 2y = -5 .......................(1)
    y 5x = 3 ..........................(2)

    from (2), we see y = 5x + 3, substitute 5x + 3 for y in (1)

    => 5x - 2(5x + 3) = -5
    => 5x - 10x - 6 = -5
    => -5x = 1
    => x = -1/5

    but y = 5x + 3
    => y = 5(-1/5) + 3
    => y = 2
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Patience View Post
    2. 4x – 12y = 5
    -x + 3y = -1
    Is there a typo here?
    4x – 12y = 5..................(1)
    -x + 3y = -1..................(2)

    from (2) we see x = 3y + 1, substitute 3y + 1 for x in (1)

    => 4(3y + 1) - 12y = 5
    => 12y + 4 - 12y = 5
    => 4 = 5
    see what i mean about the typo. One of your equations was a multiple of the other, which means we can't find a solution by this method, as there would be infinately many. I've seen a simialr thing happen with another simultaneous equations question you submitted (see number 1 here http://www.mathhelpforum.com/math-he...stitution.html ), either you made a typo in both questions, or your text book is just naughty to be giving you things like this
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