Gradeup question doubt

Question

$x(x+x'y)z(x+y+z)$ simplifies to

(A)$x+x'y$

(B)$x+y+z$

(C)$xyz$

(D)$xz$

Answer is (D)

But I can not obtain $xz$ only.

My solutions is as follows:

x(x+x'y)z(x+y+z)

=(x+y)(xz+yz+z)

=xz+xyz+xz+xyz+yz+yz

=xz+yz+xyz

=xz(1+y)+yz

=xz+yz

Can anyone tell where I made mistake?

Comments

How?

x (x + x'y) = (x+y)

Answer 1

line number 2 you are doing mistake.

x(x+x'y)=(x.x+x.x'y)=x+0=x

Comments

Thank you so much for helping me.

Answer 2

Remember few basic properties. One is

Absorption Law.  $A + AB = A ; A(A+B) = A$

And One Other Important Property is :  $A + A'B = A + B$

Apply these two Properties on the Given Boolean Equation..

$$X + X'Y = X + Y$

Now, $X(X+Y) = X$ (By Absorption Law)

And $Z(X+Y+Z) = Z$  (By Absorption Law)

Hence, We have  $X(X+X'Y)Z(X+Y+Z) = XZ$

---

Let's see where you made the mistake :

x(x+x'y)z(x+y+z)

=(x+y)(xz+yz+z)
 

See, In the 2nd line you missed $X$ because $X(X+X'Y) = X$ (Either use the Property $A+A'B=A+B$ Or Just Multiply $X$ into the parentheses ) 

Comments

Thanks. I've weak point on how to apply Absorption law.but after reading your answer I will try over come my weak point.

Answer 3

x(x+x'y)z(x+y+z)

=( x + (xx')y ) (xz + yz + z)

= x z(x + y +1) ( since xx' =0)

= xz ( since (x + y +1) = 1)

Hence xz is the answer.

Answer 4

$\color{Maroon}{x(x+\overline{x}y)z(x+y+z)}$

$= x(x+y)z(x+y+z)$     $\qquad \left [ ∵ A+\overline{A}B = A+B -- \text{By Using Redundant Literal Rule}\right ]$

$= (x.x+xy)z(x+y+z)$

$=(x+xy)z(x+y+z)$    $\qquad \left [ ∵ A.A = A \text{-- By using AND Law}\right ]$

$=xz(x+y+z)$     $\qquad \left [ ∵ A+AB = A \text{-- By using Absorption Law}\right ]$

$= x.xz+y.xz+z.xz$

$=xz+y.xz+xz$   $\qquad \left [ ∵ A.A = A \text{-- By using AND Law}\right ]$

$= xz+xz.y$   $\qquad \left [ ∵ A+A = A \text{-- By using OR Law( Identity Law)}\right ]$

$= xz(1+y)$

$=\color{Green}{xz}$    $\qquad \left [ ∵ A.(1+B) = A.1 = A \text{(1+B) = B -- By using OR Law}\right ]$

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Now I'm going to show How we actually get $\color{Purple}{A+AB = A}$

$\color{pink}{A+AB}$

$= A+A.B$

$= A(1+B)$

$=A.1$  $\qquad \left [∵ 1+B = 1 --- \text{ By using Identity Law of OR Law } \right ]$

$= \color{magenta}{A}$ $\qquad \left [∵ A.1 = A --- \text{ By using Identity Law of AND Law } \right ]$

Now, How we got this $\color{lightgreen}{A+\overline{A}B = A+B }$

$\color{lime}{A+\overline{A}B}$

$= A+\overline{A}B+0$

$= A+\overline{A}B+A.\overline{A}$  $\qquad \left [∵ A.\overline{A} = 0 --- \text{ By using AND Law } \right ]$

$= A(1+B)+\overline{A}B+A.\overline{A}$ $\qquad \left [∵ A.(1+B)= A \right ]$

$=A+AB+\overline{A}B+A.\overline{A}$

$= A.A+A.B+\overline{A}B+A.\overline{A}$  $\qquad \left [∵ A.A = A --- \text{ By using AND Law } \right ]$

$= A(A+B)+\overline{A}(A+B)$

$= (A+B)(A+\overline{A})$

$=(A+B).1$ $\qquad \left [∵ A+\overline{A} = 1 --- \text{ By using OR Law } \right ]$

$=\color{cyan}{A+B}$ $\color{Olive}{\Rightarrow\text{which is the Redundant Literal Law}}$

Comments

Nice.