The way you typed it:
2/b-2=b/b^2-3b+2 + b/2b-2 reduces to
2/b-2=1/b-3b+2 + 1/b-2
multiply each term by b
2 - 2b = 1 - 3b^2 + 2b + 1 - 2b
3b^2 - 2b = 0
b(3b-2)
b = 0 or b = 2/3, but b ≠ 0
so b = 2/3
BUT
I have a strong feeling that you meant:
2/(b-2)=b/(b^2-3b+2) + b/(2b-2)
2/(b-2) = b/((b-1)(b-2)) + b/(2(b-1))
multiply each term by 2(b-1)(b-2)
4(b-1) = 2b + b(b-2)
4b - 4 = 2b + b^2 - 2b
b^2 -4b +4 = 0
(b-2)^2 = 0
b = 2,
but b ≠2 or else we are dividing by zero in the original question
so there is no solution.
Solve rational equation
1. 2/b-2=b/b^2-3b+2 + b/2b-2
1 answer