Boolean Algebra: Simplifying $\;xyz + x'y + xyz'$ Given the following expression: $xyz + x'y + xyz'\,$ where ($'$) means complement, I tried to simplify it by first factoring out a y so I would get $\;y(xz + x' + xz').\,$ 
At this point, it appears I have several options:
A) Use two successive rounds of distributive property:
$\begin{align} y( (x + x')(z + x') + xz') ) 
&= y ( z + x' + xz')\\ & = y ( z + (x' + x)(x' + z') )\\ &= y ( z + x' + z') \\ &= y ( x') \\ &= yx'\end{align}$
B) Or I could use absorption,
$\begin{align}y ( xz + xz' + x' ) 
&= y ( x (z+z') + x') \\
& = y ( x + x' )\\
&= y ( 1) \\ 
&= y\end{align}$
I believe the second answer is correct. What am I doing wrong with option A ?

For option A, you made the error of stating $x' +1 = x'$ when its really $x' + 1 = 1$. Hence you would get the same answer for both options.
