- #1

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A'C' + ABC + AC'

C'(A'+A) + ABC

C' + ABC

or

A'C' + ABC + AC'

A'C' + A(BC + C')

A'C' + A(BC + C'(B + B'))

A'C' + A(BC + BC' + B'C'))

I can't go any further. What am I doing wrong?

The book has the answer as AB + C'

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- Thread starter success2be
- Start date

- #1

- 9

- 0

A'C' + ABC + AC'

C'(A'+A) + ABC

C' + ABC

or

A'C' + ABC + AC'

A'C' + A(BC + C')

A'C' + A(BC + C'(B + B'))

A'C' + A(BC + BC' + B'C'))

I can't go any further. What am I doing wrong?

The book has the answer as AB + C'

- #2

- 9

- 0

From the second solution

A'C' + A (BC + BC' + B'C')

A'C' + A (B (C+C') + B'C')

A'C' + A ( B + B'C' )

A'C' + A ( (B + B') (B + C') )

A'C' + A ( B + C' )

A'C' + AB + AC'

C' (A' + A) + AB

C' + AB

I wonder is it always a trial and error process in getting the solution? Wonder if I'll ever have enough time during an exam to complete.

- #3

Hurkyl

Staff Emeritus

Science Advisor

Gold Member

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Are you familiar with Karnaugh maps? (I hope I spelled that right)

- #4

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Hurkyl said:Are you familiar with Karnaugh maps? (I hope I spelled that right)

The professor have talked about the K map. The chapter I'm working on have not discussed it officially yet. It'll be the next chapter.

I'm wondering if I'll be forced to not use the K map to solve a problem on an exam. Don't know if they only care as long as the solution is correct or that the student must know every method to the solution.

I think I'll just move on to the next chapter and come back to the previous chapter question when I understand K map. Thanks.

- #5

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I'd suggest that you learn this approach well before worrying about mapping.

KM

- #6

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success2be said:

A'C' + ABC + AC'

C'(A'+A) + ABC

C' + ABC

or

A'C' + ABC + AC'

A'C' + A(BC + C')

A'C' + A(BC + C'(B + B'))

A'C' + A(BC + BC' + B'C'))

I can't go any further. What am I doing wrong?

The book has the answer as AB + C'

I am continuing on your steps but first you need to START looking at the 'and' , 'or' signs as 'plus' and 'multiply' and to start solving the problems by means of both Boolean algebra and regular algebra

A'C' + ABC + AC'

C'(A'+A) + ABC

C' + ABC

(C'+A)*(C'+B)*(C+C')

since (C+C')=1

(C'+A)*(C'+B)

(C'*C')+(C'*B)+(A*C')+(A*B)

(C'*(1+B))+(AC')+(AB)

C'+AC'+AB

C'(A+1)+AB

AB+C'

- #7

- 133

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I wonder is it always a trial and error process in getting the solution? Wonder if I'll ever have enough time during an exam to complete.

No i don't think that this is the way to think about i'd rather think about it as simplifying it as possible

I can tell that this is your first course on this,

I took this course last year and I solved your problem by just looking at it ,so don't worry you'll get better at it soon.......

- #8

- 1

- 0

A'C' + ABC + AC'

C'(A'+A) + ABC

C' + ABC

or

A'C' + ABC + AC'

A'C' + A(BC + C')

A'C' + A(BC + C'(B + B'))

A'C' + A(BC + BC' + B'C'))

I can't go any further. What am I doing wrong?

The book has the answer as AB + C'

hey u wer solving it wrnd

di like this

ur eq. wz A'C'+ABC+AC'

it will be reduced like this

C'(A'+A)+ABC

C'+ABC

NOW CONSIDER AB AS 1 TERM AND C AS SECOND

THEN (AB+C')(C+C')

C+C'=1

SO UR ANS IS AB+C'

okieeeeeeeeeee

all the bst

- #9

- 1

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that's our assignment too :)

i guess these problems are nice .. much better than electronics

i guess these problems are nice .. much better than electronics

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