Fraction and simplification 
solve: $\frac{1}{x(x-1)} + \frac{1}{x}  = \frac{1}{x-1}$
What are the possible answers ?
(A)  -1   (B) Infinitely Many Solutions (C) No solution (D) 0

The answer from where i've referred this is (B), but when i simplify it  I get (D)
My solution:
$$\frac{1}{x(x-1)} + \frac{1}{x}  = \frac{1}{x-1}$$
$$ \frac{x +x(x-1)}{x(x-1)\cdot x} = \frac{1}{x-1} \text{   (took l.c.m on l.h.s)}$$
$$ \frac{x + (x^2 -x)}{(x^2 - x)\cdot x}= \frac{1}{x-1}$$
$$\frac{x^2}{x^3 - x^2} = \frac{1}{x-1}$$
$$ x^2(x-1) = x^3 - x^2$$
$$ x^3 - x^2 = x^3 - x^2$$
Have I simplified it correctly?

You have got an identity, L.H.S. = R.H.S. which will hold for all values of X in domain of equation. 
This implies Infinite solution as domain of the equation is infinite.
In this case $X ={0,1}$ are not in domain of equation.
