[Math] 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 $

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

$=AB + C(AB)'$

$=AB+C$