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Thread: coordinates

  1. #1
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    coordinates

    http://www.emaths.co.uk/sats/2009/2009_68_Paper1.pdf

    please go to question 24 and help me answer
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  2. #2
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    Quote Originally Posted by coolubi View Post
    http://www.emaths.co.uk/sats/2009/2009_68_Paper1.pdf

    please go to question 24 and help me answer
    This looks a lot like a test, so here is just a hint:
    Notice they have told you the two equations you need to deal with. You have 2 variables and 2 equations. Do you know how to solve simultaneous equations? When you solve for x and y, make sure you pair up the right numbers with each other for the 2 points.

    Feel free to ask if you need any more help.
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  3. #3
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    Quote Originally Posted by mathemagister View Post
    This looks a lot like a test, so here is just a hint:
    Notice they have told you the two equations you need to deal with. You have 2 variables and 2 equations. Do you know how to solve simultaneous equations? When you solve for x and y, make sure you pair up the right numbers with each other for the 2 points.

    Feel free to ask if you need any more help.

    pls could you help me a little more as i am new to it(help with some working out) and only have one more week to my exam thanx
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  4. #4
    Member mathemagister's Avatar
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    Quote Originally Posted by coolubi View Post
    pls could you help me a little more as i am new to it(help with some working out) and only have one more week to my exam thanx
    The information tells you that at the points A and B, [b]both[/both] the equations $\displaystyle y=x-2$ and $\displaystyle y=x^2-2$ are correct for the same x and y values.

    So, to solve for the intercepts, you need to simultaneously solve the equations
    $\displaystyle y=x-2$ and $\displaystyle y=x^2-2$.

    You know that $\displaystyle y=x-2$, so anywhere you see a $\displaystyle y$, you are allowed to substitute it for $\displaystyle x-2$ because they are both the same thing. (Think of it like if you know $\displaystyle 3=5-2$, you can put $\displaystyle 5-2$ wherever you see a $\displaystyle 3$.

    Well, you see a $\displaystyle y$ in the second equation; you can substitute that $\displaystyle y$ for something else that equals it: $\displaystyle x-2$.

    This gives you:
    $\displaystyle x-2=x^2-2$

    $\displaystyle x=x^2$

    $\displaystyle x-x^2=0$

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

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

    to find the y value for these points simply plug them into one of the original equations:
    $\displaystyle y=x-2$
    $\displaystyle y(0) = 0 - 2 = -2$
    $\displaystyle y(1) = 1 - 2 = -1$

    These give you the points (0,-2) and (1,-1).

    Since you can see on the graph that point A lies on the y-axis (where x is always 0), you know that point A is (0,-2) and point B has to therefore be (1,-1).

    Hope that helps
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