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 ?
 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 $$y((z+z')+x')= y(1+x') = y\cdot 1 = y$$
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.
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}
A: xyz + x'y + xyz'=
multiplying (z+z') in x'y = x'y(z+z')=x'y
            z+z'=1 (do a truth table)
xyz + x'y(z+z') + xyz'
xyz + (x'yz + x'yz') + xyz'
take common 
(xyz + x'yz) +( x'yz' + xyz')
{((yz(x+x'))  +  (yz'(x+x')) }
yz+yz'
y(z+z')
:Y
