Algebra 2

4t^2-4/9(t+1) * 3t+3/2t-2

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asked by Benny
  1. Consider [4t^2-4/9(t+1)] * (3t+3)/2t-2

    3t+3/2t-2 = 3(t+1)/2(t-1)
    (4t^2-4) = 4(t+1)(t-1)

    [(4t^2-4)/9(t+1)] * (3t+3)/(2t-2)
    = {[4(t+1)(t-1)]/9(t+1)}/[3(t+1)/2(t-1)]

    Now start cancelling terms that appear in both denomimnator and numerator.

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    posted by drwls

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