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 ?

Using the distributive property (first method), we get:
$$\begin{align} xyz + x'y + xyz' & = xy(z +  z') + x'y \\ &= xy + x'y \\&= (x + x')y \\&= y\end{align}$$
You erred when you went from $ y ( z + x' + z') $ to $yx'$. You should have $$y((z+z')+x')= y(1+x') = y\cdot 1 = y$$
