Boolean Algebra: Simplify AB + A'C + B'C

I can't use K-maps. I've managed to bring it down to C + ABC', not sure if I did it right and I think I can simplify further, but don't know how to. I also feel like there's a simpler method I'm missing.

AB + A'C + B'C
= AB + A'BC + A'B'C + B'C
= AB + A'BC + B'C (1 + A')
= AB + A'BC + B'C
= ABC + ABC' + A'BC + B'C
= BC(A+A') + ABC' + B'C
= BC + B'C + ABC'
= C + ABC'

I have a second question, might as well add it here because it's also simplification of boolean algebra.

f = cx + ac'x + bc'x + a'b'c'x' (used a K-map to generate this, now I have to simplify further)
f = c'x(a+b) + cx + a'b'c'x' - no idea how to continue from here

$$AB+A'C+B'C$$

$$=AB+C(A'+B')$$

$$=AB + C(AB)'$$

$$=AB+C$$

• I would remove the word hint, sorry!! – Satish Ramanathan Oct 2 '18 at 12:47
• No harm done, feel reassured. – Yves Daoust Oct 2 '18 at 12:48
• Wow, that was easier than I thought. I got to that second line on my first try but got stuck there, I forgot about De Morgan's Laws. Thanks for the help. Could you please help with my other question (edited into my main post) if possible? – Karan Bijani Oct 2 '18 at 13:27
• Use the logic that I have used in step 3 of the first problem into your first two terms of the second problem. The last term cannot be reduced any further – Satish Ramanathan Oct 2 '18 at 13:35

\begin{aligned}AB+A'C+B'C & =AB\left(C'+C\right)+A'\left(B+B'\right)C+\left(A+A'\right)B'C\\ & =ABC'+ABC+A'BC+A'B'C+AB'C+A'B'C\\ & =ABC'+ABC+A'BC+A'B'C+AB'C\\ & =ABC'+ABC+ABC+A'BC+A'B'C+AB'C\\ & =AB\left(C'+C\right)+\left(AB+A'B+A'B'+AB'\right)C\\ & =AB+\left(A+A'\right)\left(B+B'\right)C\\ & =AB+C \end{aligned}

Make a Venn-diagram and observe that: $$(A\cap B)\cup(A^{\complement}\cap C)\cup(B^{\complement}\cap C)=(A\cap B)\cup C$$
• $c+ac'+bc'=c+a+b$ leading to $\cdots=(a+b+c)x+a'b'c'x'=(a+b+c)x+(a+b+c)'x'$ – drhab Oct 2 '18 at 14:32