How to solve this boolean algebra?

Question

I tried to solve this $(x + y) (xy'z + xyz + xy'z')$

I got $x(xz + xy'z' + yz)$ as my final result

How can I solve this?

Answer 1

$$(x + y)(xy'z+xyz+xy'z')$$

$$\implies (x + y)(xz(y'+y)+xy'z')$$

$$\implies (x + y)(xz+xy'z')$$

$$\implies xz + xy'z + xyz + xyy'z$$

$$\implies xz + xy'z + xyz$$

$$\implies xz + xz(y' + y)$$

$$\implies xz + xz$$

$$\implies xz$$

Test: Set $x = z = 1, y = $ DONTCARE. So, we set $y = 0$.

$(1 + 0)(1+0+0) = 1.1 = 1$

Answer 2

$$(x+y)*(xy'z + xyz + xy'z')$$$$ = xy'z + xyz + xy'z' + 0 + xyz + 0 $$$$= xy'z + xyz + xy'z' $$$$= x ( y'z + yz + y'z' ) $$$$= x ( z(y'+y) + y'z' )$$$$ = x$$