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I have tomorrow exam .
Faraj,
Why don't you use a Karnaugh map (K-map) to solve those problems? If we give you the answers, how is that going to help you on the exam? Where is your attempt to solve those problems? How can the first expression be solved for 3 literals when the expression contains 4 literals? Does one of the literals cancel?
Ratch
(x’y’+z)’ + z + xy + wz = xz' +yz' +z+xy+wz ; first term by DeMorgan's theorem
xz'+yz'+z+xy+wz = xz'+yz'+xy+z(1+w) = xz'+yz'+xy+z ; now only three terms
xz'+yz'+xy+z = x(y+y')z'+(x+x')yz'+xy(z+z')+(x+x')(y+y')z ;expand out the terms
= xyz'+xy'z'+xyz'+x'yz'+xyz+xyz'+xyz+xy'z+x'yz+x'y'z
=x'y'z+x'yz'+x'yz+xy'z'+xy'z+xyz'+xyz ;eliminating duplicate terms
(xy'z'+xy'z+xyz'+xyz)+(x'y'z+x'yz+xy'z+xyz)+(x'yz'+x'yz+xyz'+xyz) ;rearranging terms, we can duplicate terms as many times as we need to do so
(xy'z'+xy'z+xyz'+xyz)=xy'(z'+z)+xy(z'+z)=xy'+xy=x(y+y')=x ;reducing the first bracketed term
(x'y'z+x'yz+xy'z+xyz) = z reducing the second term is just as easy
(x'yz'+x'yz+xyz'+xyz) = y and finally the third term is reduced