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Simultaneous equations

  1. Jan 25, 2012 #1
    1. The problem statement, all variables and given/known data

    How do yo solve simultaneos equations?

    2. Relevant equations

    Am I right ?


    3. The attempt at a solution

    y = x2 – 5x + 5 question a



    5x + 3y = 30. question b



    Find location of points collide?


    Add a+b

    x2 + 2y -25=0


    quadratic formula use

    x= -2+ square root(2^2-(4x1x(-25))/2 x = 4.09901

    or

    x= -2 - square root(2^2-(4x1x(-25))/2 x=-6.09


    subsitue y etc
     
  2. jcsd
  3. Jan 25, 2012 #2

    berkeman

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    Looks like you are on the right track...
     
  4. Jan 25, 2012 #3
    Cheerio
     
  5. Jan 25, 2012 #4

    eumyang

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    I would use substitution. You already have equation a solved for y. Plug it into equation b:
    5x + 3y = 30
    5x + 3(x2 – 5x + 5) = 30
    ... and so on.
     
  6. Jan 25, 2012 #5

    NascentOxygen

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    Errrm, if eumyang is on the right track, then I'll eat my hat!

    ... and I'll post the video of it on youtube.
     
  7. Jan 25, 2012 #6

    SammyS

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    You can't use the quadratic formula on x2 + 2y -25=0, because it has two different variables in it.

    B.T.W: Yesterday, Mark44 pointed out to you that you should not use the lower case letter, x, to indicate multiplication.
     
  8. Jan 26, 2012 #7
    Hey ladies ;) + gentleman :)

    Now you have "REALLY" lost me.

    I went with eumyang- subsituations and this is what I have done.

    NB: Allright I WILL USE . FOR MULTIPLY NOW

    Ok, lets start

    1)

    y = x^2 – 5x + 5 question a



    5x + 3y = 30. question b


    Find location of points collide?



    Answer;


    5x + 3(x^2 – 5x + 5) = 30

    5x+ 15x^2 -15x+15=30

    15x^2-10x+15=30

    15x^2-10x -15 = 30

    5(3x^2 -2x -3) = 30

    use formula

    x= 2 + sqrt{40} / 6

    x= 1.3874

    or

    x=2 - sqrt{40} / 6

    x= -0.72075


    Is this the right directions?
     
  9. Jan 26, 2012 #8

    SammyS

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    The coefficient of the x2 term is 3, not 15.
    Once you get a solution for x, you should go back to an original equation to find y. To check your answer for the pair x & y, you should then plug those values into the other original equation.
     
  10. Jan 27, 2012 #9

    NascentOxygen

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    CORRECTION: I intended to write
     
  11. Jan 27, 2012 #10

    NascentOxygen

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    You must make the right hand side = 0 before using the quadratic formula.

    Taking as an example: 15x^2 - 10x - 15 = 30

    Make RHS=0:

    15x^2 - 10x - 15 - 30 = 30 - 30

    x = (10 ± sqrt(100 + 4•15•45)) / 30
     
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