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Thread: Simultaneous Equations.

  1. #1
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    Simultaneous Equations.

    I was wondering if you could doublecheck this question please:

    g) $\displaystyle y=2x$ and$\displaystyle y = 2x^2$
    y = 4 and x = 2

    How would I do this?

    h)$\displaystyle y= 6 - x^2$ and $\displaystyle y = 4x +1 $
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  2. #2
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    Re: Simultaneous Equations.

    Quote Originally Posted by JibblyJabbly View Post
    I was wondering if you could doublecheck this question please:

    g) $\displaystyle y=2x$ and$\displaystyle y = 2x^2$
    y = 4 and x = 2

    no

    How would I do this?

    h)$\displaystyle y= 6 - x^2$ and $\displaystyle y = 4x +1 $

    g)

    $\displaystyle 2x = 2x^2$

    $\displaystyle 2x - 2x^2 = 0$

    $\displaystyle 2x(1 - x) = 0$

    $\displaystyle x = 0$ and $\displaystyle x = 1$


    h)

    $\displaystyle 6-x^2 = 4x+1$

    $\displaystyle 0 = x^2 + 4x - 5$

    factor and use the zero product property to solve for x
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  3. #3
    Member sbhatnagar's Avatar
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    Thumbs up Re: Simultaneous Equations.

    In case a graph helps...

    g)$\displaystyle y=2x,y=2x^2$. The red line is $\displaystyle y=2x^2$ and the blue line is $\displaystyle y=2x$.



    Clearly solutions are $\displaystyle x=0,1$

    h)$\displaystyle y=6-x^2,y=4x+1$. The red line is $\displaystyle y=4x+1$ and the blue line is $\displaystyle y=6-x^2$.



    Clearly solutions are $\displaystyle x=1,-5$

    The value of y can be calculated just by substituting the value of x or by the graph.
    Last edited by sbhatnagar; Oct 13th 2011 at 04:32 AM. Reason: change of graph
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  4. #4
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    Re: Simultaneous Equations.

    Yet another way to do (g): $\displaystyle y=2x$ and $\displaystyle y=2x^2$:

    If x and y are not 0, $\displaystyle \frac{y}{y}= /frac{2x^2}{2x}$, [tex]1= x[/itex]. And if x= 1, what is y?

    That was "if x and y are not 0". It is easy to check that y is 0 if and only if x is and x= y= 0 satifies both equations.
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