Boolean Algebra (How to simplify)

Question

Hi! I would like to simplify a boolean expression but Im not sure if this answer is correct. Could someone simplify it? Thanks in advance! XD

 

A.) X'Y' + XY + X'Y 

Ans. X' + Y

B.)(X+Y)(X+Y')

Ans. 1

C.) X'Y + XY' +XY + X'Y'

Ans.1

D.)X'+ XY + XZ' + XY'Z'

Ans. (I don't know yet cause we just studied T5 Distributive only and I don't know how to simplify it.)

Answer 1

BASIC LAWS:

1. A(B+C) = AB + AC
2. A + A'B = A + B
3. A + AB = A
4. A(A+B) = A
5. A(A'+B) = AB

 

A.) X'Y' + XY + X'Y  

= X'Y' + Y(X+X')      [LAW 1]

= X'Y' + Y(1)

= Y + X'                   [LAW 2]

B.)(X+Y)(X+Y')

= XX+XY'+YX+YY'     [LAW 1]

= 1+X(Y'+Y)+0

= 1+X(1)

= 1+X

= 1

C.) X'Y + XY' +XY + X'Y'
= X'Y + X'Y' + XY' + XY      [REGROUP TERMS]

= X'(Y+Y') + X(Y'+Y)         [LAW 1]

= X'(1) + X(1)

= X' + X

= 1

D.)X'+ XY + XZ' + XY'Z'

= X'+Y + XZ'(1+Y')           [LAW 1]

= X' + Y + XZ'(1)

= X' + Y + XZ'

= X' + XZ' + Y                [REGROUP TERMS]

= X' + Z' + Y                  [LAW 2]