# 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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