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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. $1 \ 0 \ 0 \ 0$
asked in Digital Logic by Veteran (59.6k points)
edited by | 1.4k views
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.
0
In both GO PDF and Hardcopy, all equations are in single line.
0
For those who are wondering, 1 + 0 + 1 = 0, its not addition, it('+') is logical OR operation.
0
@chotu please provide your solution

4 Answers

+19 votes
Best answer
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
answered by Boss (45.3k points)
edited by
0

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.

+8 votes
Answer: C
answered by Boss (34k points)
0
x + y + z =  1 + 0 + 1 = 0 ? only right
+1
Got it. basic OR gate properties
+2 votes
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 is C
answered by (459 points)
+1 vote
Answer -    C. 1 0 1 1   * think without ink*
answered by Active (3.9k points)
Answer:

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