algebra2

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(3b-2)/b+1=4-(b+2/b-1)
please help

  • algebra2 -

    Multiply both sides by b. (I am assuming that (b+1) and (b-1) are not denominators.)

    3b-2+b = 4b-b(b+2/b-1)

    3b-2+b = 4b-b^2-2+b

    (b^2 = b squared.)

    Combine like terms.

    2b-2 = 5b-b^2-2

    Add 2 to both sides.

    2b = 5b-b^2

    Bring all terms to the left side of the equation.

    b^2-3b = 0

    Factor.

    b(b-3) = 0

    Therefore b = 3 or 0.

    I hope this helps. Thanks for asking.

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