Boolean Algebra Minimization

Question

Prove that

$\bar{A}B + AC + BC = \bar{A}B + AC$

with the help of boolean algebraic manipulations. I have no clue from where to start, how should I tackle these type of questions?

Or

$$ \left(\neg A \wedge B\right)\vee \left(A \wedge C\right)\vee\left(B \wedge C\right) = \left(\neg A \wedge B\right)\vee \left(A \wedge C\right)$$

Answer 1

HINT Since you want to get rid of the term $BC$, maybe it makes sense to do something to it : $$\begin{align}\bar{A}B + AC + BC &= \bar{A}B + AC + (A+\bar{A})BC\\&=\bar{A}B + AC + ABC+\bar{A}BC\end{align} $$

Group first and last terms
Group middle two terms

Answer 2

Just to start from the obvious, if $A$ is false, then clearly either $B$ or $B$ and $C$ must be true. Right away, this means that when $A$ is false, $B$ cannot be false. So this suggests $\neg A \wedge B$. Now, if $A$ is true then either $C$ is true or $C$ and $B$ is true--either way $C$ is true, therefore it's clear that this is equivalent to $(A \wedge B)\vee (\neg A \wedge C)$.