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algebraII

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3b-2/b+1=4-b+2/b-1
please help me I'm stuck

  • algebraII -

    retype with brackets to establish the correct order of operation.

    did you mean
    (3b-2)/b+1 = (4-b+2)/(b-1) or
    3b - 2/(b+1) = 4-b + 2/b-1 or ...

  • algebraII -

    3b-2 b+2
    ____=4-_____
    b+1 b-1

  • algebraII -

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

  • algebraII -

    still not totally clear, I will assume you mean
    (3b-2)/(b+1)=4-(b+2)/(b-1)

    so the LCD is (b-1)(b-1), let's multiply each term by that

    (3b-2)(b-1) = 4(b-1)(b+1) - (b+2)(b+1)
    for which I go
    b = 4

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