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josh3012 · Sep 22, 2014, 07:01 AM

Using Boolean algebra;

1. (X + Y + Z)(X + Y + Z')
2. A'BC' + ABC' + BC'D + BC + B'
3. A'C + ABC' + A'B + AB'
4. AB' + B'C' + AB + B'C
My answer

1. X + Y.
2. C'D + 1
3. (Somehow I cant simplify this)
4. B' + A
Could anyone confirm this is correct and how do I simplify Q.3?

josh3012 · Sep 22, 2014, 07:14 AM

Well, I did manage to simplify for Q.3. My answer is AB + AC + ABC'. And I don't know if it can be simplify again, but this is so far I can get.

josh3012 · Sep 22, 2014, 09:03 AM

sorry, correction for Q2. My answer is BC'D + A.

I did recheck my answer for Q.3, it's wrong.

This is my working for Q.3,

A'C = A(A'+C) = AC.
A'B = A(A'+B) = AB.
B'A = B(B'+A) = AB or BA.

AC + AB + AB + ABC'
AC + AB + ABC'. <------but I verify this with truth table, its wrong. Help! I want to know what's wrong.

InfoJunkie4Life · Nov 17, 2014, 07:25 PM

Just Curious here, does
$ A' = \bar{A} $
$ B' = \bar{B} $
$ C' = \bar{C} $
If so,
$ A\bar{B} + \bar{B}\bar{C} + AB + \bar{B}C = $
$ \bar{B}(A + \bar{C} + C) + AB $
$ \bar{B}(A + 1) + AB $
$ \bar{B}(1) + AB $
$ \bar{B} + AB $

Excuse me if this makes no sense, this is my first dabble on boolean math. Reviewing the identities I think it makes sense...I'm not sure how to carry it from there...