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

Question

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

Answer 1

$AB+A'C+B'C $

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

$=AB + C(AB)'$

$=AB+C$

Answer 2

$\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} $

Constructive advice:

Make a Venn-diagram and observe that: $$(A\cap B)\cup(A^{\complement}\cap C)\cup(B^{\complement}\cap C)=(A\cap B)\cup C$$