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 ?

Second answer is correct as can also be done as follows
\begin{align}
xyz+x^{\prime}y+xyz^{\prime}&=xyz+xyz^{\prime}+x^{\prime}y\qquad\text{commutative law}\\&=xy(z+z^{\prime})+x^{\prime}y\\&=xy.1+x^{\prime}y\\&=(x+x^{\prime})y=1.y=y
\end{align}
While as in A) Note that 
\begin{align}
y((x+x^{\prime})(z+x^{\prime})+xz^{\prime})&=y(z+x^{\prime}+xz^{\prime})\\&=yz+x^{\prime}y+xyz^{\prime}\neq xyz+x^{\prime}y+xyz^{\prime}
\end{align}
