Boolean equation simplification

Question

This is the problem: XY’ + XYZ + XY'Z= X + Y'Z

And so far I have this,

XY’ + XYZ + XY'Z= X + Y'Z
X(Y’ + YZ + Y’Z) Factor out X
X(Y’ + Z + Y’Z) De Morgan

Any tips on how to proceed? I know it shouldn't be very hard but I am stuck..

EDITED. How about this:

This is the problem: (AB’ + ABC)’ +A’BC’= A’ + BC’ And so far I have this,

  A’B + A’B’C’ + A’BC’      
  A’(B + B’C’+ BC’)  
  A’(B + (B’+B)C’)  
      ???

Any tips on how to proceed? I know it shouldn't be very hard but I am stuck..

Answer 1

You incorrectly applied DeMorgan's (there's no need for DeMorgan's).

After factoring out $X$, you have $$\begin{align} X(Y' + YZ +Y'Z) & = X[Y' + (\underbrace{Y+Y'}_{\large =\,1})Z] \\ \\ & = X(Y' + Z)\end{align}$$

as desired.

ADDED after edited question:

You didn't apply DeMorgan's correctly in your first "move". Starting from the beginning:

$$\begin{align} (AB’ + ABC)’ +A’BC’ & = (AB')'(ABC)' + A'BC' \\ \\ & = (A'+B)(A'+B'+C') + A'BC' \\ \\ &= [A' + (B(B'+C'))] + A'BC' \\ \\ & = A' + BC' + A'BC'\\ \\ &= A' + BC'\end{align}$$