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