Exercise regarding boolean algebra?

Question

We need to simplify

1. $AC+A'B'C$
2. $Y=A'B' +A'B C'+(A+C')'$

For (1) I wrote $C(A+A'B')$ but the result must be $AC+ B'C$. How do I get that to happen?

I tried to simplify (2) using deMorgan but no results. The result must be $Y=A'$ but I cant get it as much as I try

Answer 1

In both cases you should use the identity, that $X=XY+XY'$

1. $\ldots =C(A+A'B') = C(A + A + A'B') = C(\underbrace{A+(AB}_{=A}+\underbrace{AB')+A'B'}_{=B'}) = C(A+B') = AC+ B'C$
2. $A'B'+A'BC'+(A+C')' = A'B'+A'BC'+A'C = A'(B'+BC'+C)$ $= A'(\underbrace{(B'C+B'C')+BC'+(BC+B'C)}_{=1}) = A'$

Answer 2

For $(1)$, we can use two forms of the distributive law - one you used, as well as: $$X + YZ = (X + Y)(X+Z)$$

$$\begin{align} AC + A'B'C = C(A + A'B') & = C\Big((A+A')(A+B')\Big) \\ \\ & = C\Big((1)(A+B')\Big) \\ \\ & = C(A+B') \\ \\ &= AC + B'C\end{align}$$

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For $(2)$ we can use the identity given by: $X = X(1) = X(\underbrace{Y + Y'}_{\large =\, 1}) = XY + XY'\tag{ID}$:

$$\begin{align} A'B'+A'BC'+(A+C')' & = A'B'+A'BC'+A'C \\ \\ & = A'(B'+BC'+C)\\ \\ & = A'\Big((B'C + B'C') + BC' + (BC + B'C)\Big) \tag{ID}\\ \\ & = A'\Big(\underbrace{BC + B'C + BC' +B'C'}_{\large = 1}\Big)\\ \\ & = A' \end{align}$$

In the second to last line, note that one of $BC, B'C, BC', B'C'$ is necessarily true (since there are four possible truth value assignments to two variables).