I need help starting this, I don't want a plain answer just a what to do.
(bc'+a'd)(ab'+cd')-Simplify
My best guess is to use the distributive
property.
(a+b)(a+c)= a(b+c). But now that there is a 4th variable what do I do?
use it twice
(bc'+a'd)(ab'+cd')
= bc'(ab'+cd') + a'd(ab'+cd')
= abb'c' + bcc'd' + aa'b'd + a'cdd'
Now, since xx' = 0, the result is 0
OOK, thank you, I see how you did that step. that makes perfect sense.
so simplified you get
ac'+bd'+b'd+a'c
no, the result is 0, since
abb'c = a(bb')c = a0c = 0
each term is 0
To simplify the given expression (bc'+a'd)(ab'+cd'), you are correct in thinking of using the distributive property. The distributive property states that for any three variables a, b, and c, (a + b)(c) = ac + bc.
However, in this case, we have four variables: a, b, c, and d. This means we need to apply the distributive property twice, taking each term in the first parentheses and multiplying it with each term in the second parentheses. Let's break it down step by step:
First, distribute the first term (bc') with the terms inside the second parentheses (ab' + cd'):
(bc') * (ab' + cd') = (bc')(ab') + (bc')(cd')
Next, distribute the second term (a'd) with the terms inside the second parentheses (ab' + cd'):
(a'd) * (ab' + cd') = (a'd)(ab') + (a'd)(cd')
Now, you can simplify each of the resulting terms individually:
(bc')(ab') = bca'b'
(bc')(cd') = bcc'd'
(a'd)(ab') = a'dab'
(a'd)(cd') = a'dcd'
Finally, combine the simplified terms:
bca'b' + bcc'd' + a'dab' + a'dcd'
Now, you have simplified the original expression.