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The simultaneous equations on the Boolean variables $x, y, z$ and $w$,

• $x + y + z = 1$
• $xy = 0$
• $xz + w = 1$
• $xy + \bar{z}\bar{w} = 0$

have the following solution for $x, y, z$ and $w,$ respectively:

1. $0 \ 1 \ 0 \ 0$
2. $1 \ 1 \ 0 \ 1$
3. $1 \ 0 \ 1 \ 1$
4. $1 \ 0 \ 0 \ 0$

Although option rejection is good approach. But 4 variable k-map could be used to find all solutions(means cells where one will come for all expression) like 1010 is also a solution.
In both GO PDF and Hardcopy, all equations are in single line.
For those who are wondering, 1 + 0 + 1 = 0, its not addition, it('+') is logical OR operation.
Checking using options

x+y+z=1 , so anyone of x,y,z must be 1, cannot eliminate any option.

xy=0 => B eliminated as x,y both 1

xz+w=1 => either x,z both 1 or w is 1, A,D => neither true

So C is correct.
In this type of problems checking option can give you result faster  !

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Take each option one by one and try to put the values of $x, y, z$ and $w$ in question:
1.

• $0+1+0 =1$
• $0.1 =0$
• $0.0+0 =0$  (went wrong)      So, this is not the right option

2.

• $1+1+0 =1$
• $1.1 =1$ (went wrong)          not right

3 .

• $1+0+1 =1$
• $1.0 =0$
• $1.1+1=1$
• $1.0 +1.0 =0$     This is the right option

Correct Answer: $C$

### 1 comment

Look at each equation and see which option is not satisfying that equation. That option will get eliminated.
Equation 1 is satisfied by all options.
Equation 2 eliminate option B
Equation 3 eliminate options A and D.

x + y + z =  1 + 0 + 1 = 0 ? only right
Got it. basic OR gate properties
My approach option elimination:-
as xy=0
then x= or y= 0 or x and y both are 0. So B is eliminated.

xy+z'w'=0
implies these xy=0 and z'w' =0
implies these either z'=0 or w'=0 or both
impiles these either z=1 or w=1 or both.
So A and D are eliminated.

Answer -    C. 1 0 1 1   * think without ink*

option c is right.