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Thread: simultaneous equation help

  1. #1
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    simultaneous equation help

    i'd be really grateful if someone could complete this simultaneous equation with working out explaining each stage of the process. thanks.

    y = 2x + 3

    x2 + y2 = 2
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  2. #2
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    I assume that the two equations are:

    $\displaystyle y = 2x + 3$ and $\displaystyle x^2 + y^2 = 2$.

    Substituting into equation 2 gives

    $\displaystyle x^2 + (2x + 3)^2 = 2$.

    Now expand, collect like terms, and solve.
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  3. #3
    Senior Member eumyang's Avatar
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    Quote Originally Posted by tylerdurden View Post
    i'd be really grateful if someone could complete this simultaneous equation with working out explaining each stage of the process. thanks.

    y = 2x + 3

    x2 + y2 = 2
    You mean this?
    $\displaystyle y\,=\,2x\,+\,3$
    $\displaystyle x^2\,+\,y^2\,=\,2$

    Plug the first equation into the second where you see the y:
    $\displaystyle x^2 + (2x + 3)^2 = 2$

    Solve for x:
    $\displaystyle x^2 + 4x^2 + 12x + 9 = 2$
    $\displaystyle 5x^2 + 12x + 7 = 0$
    $\displaystyle (x + 1)(5x + 7) = 0$

    Set each factor equal to 0 and you'll get
    $\displaystyle x = -1$ or $\displaystyle x = {-\frac{7}{5}}$

    Plug both values into the first equation to get the corresponding y values.


    EDIT: Beaten to it!
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  4. #4
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    cheers guys. appreciate the help.

    the only bit i'm not too sure on is how to get from

    x2 + (2x+3)2 = 2 to

    x2 + 4x2 + 12x = 9 =2

    would really appreciate if you could explain what i have to do.


    thanks.
    Last edited by tylerdurden; Jul 3rd 2010 at 07:01 AM.
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  5. #5
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    Remember that $\displaystyle (2x + 3)^2 = (2x + 3)(2x + 3)$.

    Expand it out.
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  6. #6
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    4x2 + 6x + 6x + 9

    thanks for the help!
    Last edited by tylerdurden; Jul 3rd 2010 at 07:17 AM.
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  7. #7
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    not quite sure how you get from

    (x+1) (5x +7) = 0

    to

    x= -1 y = 7/5

    and why

    thanks
    Last edited by tylerdurden; Jul 3rd 2010 at 08:19 AM.
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  8. #8
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    When you have two numbers being multiplied together to give $\displaystyle 0$, that means at least one of them has to be $\displaystyle 0$.

    So if $\displaystyle (x + 1)(5x + 7) = 0$ then $\displaystyle x + 1 = 0$ or $\displaystyle 5x+ 7 = 0$.
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