Help Me simplify: C*(A+B) + ~A*B












2












$begingroup$


I know the answer is AC + ~AB, but how?



I have tried:



B*(~A+C) + A*C and stop.



Also, I have tried:



AC + BC + ~AB and have stopped.



There seems nowhere to go in either case.






digital-logic electrical







share|improve this question









New contributor




Alan Kazemian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • 3




    $begingroup$
    Show what you've tried and be explicit with where you are confused. Help us to help you.
    $endgroup$
    – Bort
    5 hours ago
















2












$begingroup$


I know the answer is AC + ~AB, but how?



I have tried:



B*(~A+C) + A*C and stop.



Also, I have tried:



AC + BC + ~AB and have stopped.



There seems nowhere to go in either case.






digital-logic electrical







share|improve this question









New contributor




Alan Kazemian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 3




    $begingroup$
    Show what you've tried and be explicit with where you are confused. Help us to help you.
    $endgroup$
    – Bort
    5 hours ago














2












2








2





$begingroup$


I know the answer is AC + ~AB, but how?



I have tried:



B*(~A+C) + A*C and stop.



Also, I have tried:



AC + BC + ~AB and have stopped.



There seems nowhere to go in either case.






digital-logic electrical







share|improve this question









New contributor




Alan Kazemian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




I know the answer is AC + ~AB, but how?



I have tried:



B*(~A+C) + A*C and stop.



Also, I have tried:



AC + BC + ~AB and have stopped.



There seems nowhere to go in either case.






digital-logic electrical




digital-logic electrical






share|improve this question









New contributor




Alan Kazemian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question









New contributor




Alan Kazemian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




share|improve this question








edited 4 hours ago









pipe

10.1k42656




10.1k42656






New contributor




Alan Kazemian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 5 hours ago









Alan KazemianAlan Kazemian

183




183




New contributor




Alan Kazemian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Alan Kazemian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Alan Kazemian is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 3




    $begingroup$
    Show what you've tried and be explicit with where you are confused. Help us to help you.
    $endgroup$
    – Bort
    5 hours ago














  • 3




    $begingroup$
    Show what you've tried and be explicit with where you are confused. Help us to help you.
    $endgroup$
    – Bort
    5 hours ago








3




3




$begingroup$
Show what you've tried and be explicit with where you are confused. Help us to help you.
$endgroup$
– Bort
5 hours ago




$begingroup$
Show what you've tried and be explicit with where you are confused. Help us to help you.
$endgroup$
– Bort
5 hours ago










1 Answer
1






active

oldest

votes


















7












$begingroup$

I will not give you the full solution, but the required non-trivial sidestep. You have got this far:



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


Now here is the sidestep. We know that (A+A') is 1, so we can do:



 = AC + (A + A')BC + A'B


From here you will need to expand it and use the "OR absorption law" twice.






share|improve this answer











$endgroup$













  • $begingroup$
    @AlanKazemian (Be sure and hit the check mark so you can close the question)
    $endgroup$
    – KingDuken
    4 hours ago










  • $begingroup$
    @Eugene Sh. Expanding the brackets was an obvious step but your "sidestep" was one that I had not considered. Very useful and I will bear this in mind for future tasks. A Karnaugh Map would also have provided a easy solution to the problem.
    $endgroup$
    – Pzy
    2 hours ago











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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7












$begingroup$

I will not give you the full solution, but the required non-trivial sidestep. You have got this far:



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


Now here is the sidestep. We know that (A+A') is 1, so we can do:



 = AC + (A + A')BC + A'B


From here you will need to expand it and use the "OR absorption law" twice.






share|improve this answer











$endgroup$













  • $begingroup$
    @AlanKazemian (Be sure and hit the check mark so you can close the question)
    $endgroup$
    – KingDuken
    4 hours ago










  • $begingroup$
    @Eugene Sh. Expanding the brackets was an obvious step but your "sidestep" was one that I had not considered. Very useful and I will bear this in mind for future tasks. A Karnaugh Map would also have provided a easy solution to the problem.
    $endgroup$
    – Pzy
    2 hours ago
















7












$begingroup$

I will not give you the full solution, but the required non-trivial sidestep. You have got this far:



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


Now here is the sidestep. We know that (A+A') is 1, so we can do:



 = AC + (A + A')BC + A'B


From here you will need to expand it and use the "OR absorption law" twice.






share|improve this answer











$endgroup$













  • $begingroup$
    @AlanKazemian (Be sure and hit the check mark so you can close the question)
    $endgroup$
    – KingDuken
    4 hours ago










  • $begingroup$
    @Eugene Sh. Expanding the brackets was an obvious step but your "sidestep" was one that I had not considered. Very useful and I will bear this in mind for future tasks. A Karnaugh Map would also have provided a easy solution to the problem.
    $endgroup$
    – Pzy
    2 hours ago














7












7








7





$begingroup$

I will not give you the full solution, but the required non-trivial sidestep. You have got this far:



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


Now here is the sidestep. We know that (A+A') is 1, so we can do:



 = AC + (A + A')BC + A'B


From here you will need to expand it and use the "OR absorption law" twice.






share|improve this answer











$endgroup$



I will not give you the full solution, but the required non-trivial sidestep. You have got this far:



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


Now here is the sidestep. We know that (A+A') is 1, so we can do:



 = AC + (A + A')BC + A'B


From here you will need to expand it and use the "OR absorption law" twice.







share|improve this answer














share|improve this answer



share|improve this answer








edited 2 hours ago

























answered 5 hours ago









Eugene Sh.Eugene Sh.

7,5581830




7,5581830












  • $begingroup$
    @AlanKazemian (Be sure and hit the check mark so you can close the question)
    $endgroup$
    – KingDuken
    4 hours ago










  • $begingroup$
    @Eugene Sh. Expanding the brackets was an obvious step but your "sidestep" was one that I had not considered. Very useful and I will bear this in mind for future tasks. A Karnaugh Map would also have provided a easy solution to the problem.
    $endgroup$
    – Pzy
    2 hours ago


















  • $begingroup$
    @AlanKazemian (Be sure and hit the check mark so you can close the question)
    $endgroup$
    – KingDuken
    4 hours ago










  • $begingroup$
    @Eugene Sh. Expanding the brackets was an obvious step but your "sidestep" was one that I had not considered. Very useful and I will bear this in mind for future tasks. A Karnaugh Map would also have provided a easy solution to the problem.
    $endgroup$
    – Pzy
    2 hours ago
















$begingroup$
@AlanKazemian (Be sure and hit the check mark so you can close the question)
$endgroup$
– KingDuken
4 hours ago




$begingroup$
@AlanKazemian (Be sure and hit the check mark so you can close the question)
$endgroup$
– KingDuken
4 hours ago












$begingroup$
@Eugene Sh. Expanding the brackets was an obvious step but your "sidestep" was one that I had not considered. Very useful and I will bear this in mind for future tasks. A Karnaugh Map would also have provided a easy solution to the problem.
$endgroup$
– Pzy
2 hours ago




$begingroup$
@Eugene Sh. Expanding the brackets was an obvious step but your "sidestep" was one that I had not considered. Very useful and I will bear this in mind for future tasks. A Karnaugh Map would also have provided a easy solution to the problem.
$endgroup$
– Pzy
2 hours ago










Alan Kazemian is a new contributor. Be nice, and check out our Code of Conduct.










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