Digital Electronics - Boolean Algebra and Logic Simplification - Discussion
Discussion Forum : Boolean Algebra and Logic Simplification - General Questions (Q.No. 18)
18.
AC + ABC = AC
Discussion:
12 comments Page 1 of 2.
Raj said:
3 years ago
Question: AC+ABC = AC
Sol:-
AC+ABC = AC
Taking AC common in LHS.
AC(1+B) = AC,
AC(1) = AC,
i.e,[1+B=1]
AC = AC,
HENCE PROVED.
Sol:-
AC+ABC = AC
Taking AC common in LHS.
AC(1+B) = AC,
AC(1) = AC,
i.e,[1+B=1]
AC = AC,
HENCE PROVED.
(1)
PRIYANKA said:
1 decade ago
given expression is AC+ABC=AC
L.H.S
by associative law AC(1+B)
And we know that 1+A=1
.
. . AC(1)
and we also know that A.1=A
.
. . L.H.S=AC
THUS L.H.S.= R.H.S.
L.H.S
by associative law AC(1+B)
And we know that 1+A=1
.
. . AC(1)
and we also know that A.1=A
.
. . L.H.S=AC
THUS L.H.S.= R.H.S.
STUDENT OF PVPIT said:
1 decade ago
AC+ABC.
= AC(B+B')+ABC.....B+B'=1.
= ABC+AB'C+ABC.
= ABC+AB'C......A+A=A.
= AC(B+B').
= AC.....B+B'=1.
= AC(B+B')+ABC.....B+B'=1.
= ABC+AB'C+ABC.
= ABC+AB'C......A+A=A.
= AC(B+B').
= AC.....B+B'=1.
Rohan. said:
1 decade ago
AC+ABC.
Taking AC common,
AC(1+B).
Here 1+B = 1.
Hence answer is AC.
Taking AC common,
AC(1+B).
Here 1+B = 1.
Hence answer is AC.
_Meth_ said:
6 years ago
AC + ABC
= AC(1+B) : Redundancy law.
= AC : Identity law.
= AC(1+B) : Redundancy law.
= AC : Identity law.
Sanjeev said:
1 decade ago
AC+ABC
Taking AC as common
AC(1+C) =AC (Since 1+x=x)
Taking AC as common
AC(1+C) =AC (Since 1+x=x)
Usman kabir suleiman said:
6 years ago
AC+ABC = AC.
= AC(1+B).
= 1+B = 1.
AC(1) = AC.
= AC(1+B).
= 1+B = 1.
AC(1) = AC.
(1)
AWEL said:
10 years ago
AC+ABC.
= A(C+BC)....X+XY = X.
= A(C) = AC.
= A(C+BC)....X+XY = X.
= A(C) = AC.
Ifan said:
5 years ago
I don't understand. Please, anyone, explain.
Inder said:
1 decade ago
AC+ABC=AC
AC(1+B)=AC
AC=AC ...(1+x=1)
AC(1+B)=AC
AC=AC ...(1+x=1)
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