Boolean algebra: $(x+y)(x’+z)(y+z) = (x+y)(x’+z)$

Question

Could someone explain to me how this simplification is derived?

$(x+y)(x’+z)(y+z) = (x+y)(x’+z)$

Answer 1

$\begin{align} (x + y)(x' + z)(y + z) & = (xz + x'y + yz)(y + z)\\ & = (xyz + xz + x'y + x'yz + yz + yz)\\ & = [xz(1 + y) + x'y(1 + z) + yz]\\ & = (xz + x'y + yz)\\ & = (x + y)(x' + z)\end{align}$