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Math Help - Help! Solving simultaneous equations using the graph method

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    Help! Solving simultaneous equations using the graph method

    Hi all, I would be very grateful if someone could assist me with solving simultaneous equations using the graph method.

    Here is the question:
    Robot A follows a route which has the equation y = x - 2 and Robot B follows a route which has the equation 2y = x + 7. Find the co-ordinates that both robots pass over by plotting the two routes.


    Many thanks in advance.
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    Quote Originally Posted by IPFreely View Post
    Hi all, I would be very grateful if someone could assist me with solving simultaneous equations using the graph method.

    Here is the question:
    Robot A follows a route which has the equation y = x - 2 and Robot B follows a route which has the equation 2y = x + 7. Find the co-ordinates that both robots pass over by plotting the two routes.
    On the graph do the following
    -take one value of x , find the corresponding value of y in the first equation ,(x,y) is one point on that path

    -Repeat above steps three times and join( and extend) the three points this is the first path

    -Do the same thing for next equation and march ahead
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    Quote Originally Posted by IPFreely View Post
    Hi all, I would be very grateful if someone could assist me with solving simultaneous equations using the graph method.

    Here is the question:
    Robot A follows a route which has the equation y = x - 2 and Robot B follows a route which has the equation 2y = x + 7. Find the co-ordinates that both robots pass over by plotting the two routes.


    Many thanks in advance.
    well as you know we have to line then up.
    so....

    y=x-2..... (we have to bring numbers to one side and the un-known's to 2y=x+7 the other)

    y-x=-2
    2y-x=7..... we have to change the signs so the x's well cancel

    -y+x=2
    2y-x=7 as you can see the x's well go

    y=9

    now you fill in for y.

    y=x-2
    9=x-2
    9+2=x
    11=x

    do you understand it? any questions just ask
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