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23 November, 11:34

Solve (3/x-3) = (x/x-3) - (3/2) for x and determine if the solution is extraneous or not

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  1. 23 November, 12:35
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    3 / (x-3) = x / (x-3) - 3/2

    6/2 (x-3) = 2x/2 (x-3) - 3 (x-3) / 2 (x-3)

    6=2x-3 (x-3)

    6=2x-3x+9

    x=3

    x=3 makes the denominator x-3 a zero, so the solution is extraneous,
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