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Math Help - solving 2 linear equations

  1. #1
    Super Member bigwave's Avatar
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    solving 2 linear equations

    this should be easy but keep getting different answers every time I do it.

    2x + \frac{5}{y} = -11

    4x-\frac{3}{y}=\frac{21}{3}

    multiplied eq1 by 3 and eq2 by 5 to get

    6x+\frac{15}{y} = -33
    20x-\frac{15}{y}=\frac{105}{3}

    adding gives

    26x = -33 + \frac{105}{3}

    multiply thru by 3

    78x = -99 + 105 = 6

    x = \frac{6}{78}

    before deriving y using x it is hard to believe that this value is correct??
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  2. #2
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by bigwave View Post
    this should be easy but keep getting different answers every time I do it.

    2x + \frac{5}{y} = -11

    4x-\frac{3}{y}=\frac{21}{3}

    multiplied eq1 by 3 and eq2 by 5 to get

    6x+\frac{15}{y} = -33
    20x-\frac{15}{y}=\frac{105}{3}

    adding gives

    26x = -33 + \frac{105}{3}

    multiply thru by 3

    78x = -99 + 105 = 6

    x = \frac{6}{78}

    before deriving y using x it is hard to believe that this value is correct??
     x = \frac {1}{13} \;\; y = - \frac {13}{29}


    so your solution is correct

    now just put that x in any of the first 2 equations and you will get y



    Edit: perhaps it would be easier to multiply first one with -2 and add them ... but it's the same result
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  3. #3
    Super Member bigwave's Avatar
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    ok, just to finish it out

    x=\frac{6}{78}=\frac{1}{13}

    2(\frac{1}{13}) + \frac{5}{y} = -11

    2y + 65 = -143y

    145y = -65

    y = \frac{-65}{145} = -\frac{13}{29}
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  4. #4
    Senior Member yeKciM's Avatar
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    yes you have done it correctly
    always look for the simpler way to solve , so there will be less room for the error
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  5. #5
    Super Member bigwave's Avatar
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    actually, it was sugested that when fractions show in simultanious equations that it is best to multiply or cancel them out as an easy way to do it... however I see your point in that -2 would of left the fractions but the x would of canceled... quess the y would not of been that hard to get... yes i always seem to introduce error with a lot of steps
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  6. #6
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    One "difficulty", by the way, is that these are not "linear equations"!
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  7. #7
    Flow Master
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    Quote Originally Posted by HallsofIvy View Post
    One "difficulty", by the way, is that these are not "linear equations"!
    Hah! I initially thought the same thing. But then I decided that one could reasonably argue that they are linear in the variables x and 1/y .....
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  8. #8
    Super Member bigwave's Avatar
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    when I ploted them they were not lines

    yep, you are right, when I ploted them they were not lines. I presume this is because the y is in the denominator making it rational resulting a graph with an asymptotes.
    I did try to make this into the y=mx+b but of course cannot.
    however the text did say they were linear equations but doesn't look like it.
    interesting catch not sure where the red flag was.
    Last edited by bigwave; September 23rd 2010 at 11:47 AM. Reason: grammar
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