I am very new to the circuit world, having just learned it recently in school. The teachers have given us some homework on the simplification of Boolean algebra, although there were no answers so it's very frustrating not knowing whether I've done them right or wrong.

Right now, I have a few practice questions where we have to simplify to the simplest form that I'm not too sure whether I did right or wrong. It'd be great if you could tell me how I did on these questions.

*underline means that it's a NOTed*

(A + B) (AB + C) C
= (A + B) C
= AC + BC

(A + B) C + (AB + C)
= AC + BC + AB + C
= AC + BC + AB

 

The first one seems correct. You noticed that
(AB'+C)C=AB'C+C=C(1+AB')=C(1)=C

Why didn't you do the same with B'C+C in the second question?

 

Oh I see, so it should be like this?

(A + B') C + (AB' + C)
= AC + B'C + AB' + C
= AC + AB' + (B'C + C)
= AC + AB' + C (1 + B')
= AC + AB' + C
= AB' + (AC + C)
= AB' + C(1 + A)
= AB' + C?

 

Give this tutorial a glance. There are a few identities explained that you should know to help you break this down.
http://www.allaboutcircuits.com/vol_4/chpt_7/5.html

Sorry, the first one does look right. Second one looks good on your second attempt!

 

I was under the impression that I'd have to simply follow the laws of Boolean algebra and sub them into each equation, now I'm slowly seeing why these equations work. That was the sole reason I think I even got the first one right ><

 

@Frank2368

Yes, this is what I was talking about.

Boolean laws form the basis, but identities such as the ones in the link make your life much easier.