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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
edited | 1.2k views
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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.

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
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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.

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x + y + z =  1 + 0 + 1 = 0 ? only right
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Got it. basic OR gate properties
Answer -    C. 1 0 1 1   * think without ink*
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.