word problem, please help

In 1994 the moose population in a park was measured to be 6400. By 1996, the population was measured again to be 6800. If the population continues to change linearly:

Find an equation for the moose population, y,in terms of x, the years since 1994.

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  1. The two-point form of a straight line passing through the points ( x1, y1 ) and ( x2 , y2 ) is given by:

    y − y1 = ( y2 − y1 ) * ( x − x 1 ) / ( x2 − x1 )

    Ini this case:

    x1 = 1994 , x2 = 1996

    y1 = 6400 , y2 = 6800

    So:

    y − y1 = ( y2 − y1 ) * ( x − x 1 ) / ( x2 − x1 )

    y − 6400 = ( 6800 − 6400 ) * ( x − 1994 ) / ( 1996 − 1994 )

    y − 6400 = 400 * ( x − 1994 ) / 2

    y − 6400 = 200 * ( x − 1994 )

    y − 6400 = 200 * x − 200 * 1994

    y − 6400 = 200 x − 398800 Add 6400 to both sides

    y − 6400 + 6400 = 200 x − 398800 + 6400

    y = 200 x - 392400

    Proof:

    x = 1994

    y = 200 x - 392400 = 200 * 1994 - 392400 = 398800 - 392400 = 6400

    x = 1996

    y = 200 x - 392400 = 200 * 1996 - 392400 = 399200 - 392400 = 6800

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  2. thank you

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  3. Now it is asking what will be the prediction for 2006?

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