Hello,
Im trying to solve this with identities



xz+z /z + zxy = z
z ( x +/x + xy) <-- x +/x = 1 (identity)
z (xy) = z so now what? Im trying to find identity to get rid xy but I dont know how to do it ..

 

Hello,
Im trying to solve this with identities

View attachment 51689

xz+z /z + zxy = z
z ( x +/x + xy) <-- x +/x = 1 (identity)
z (xy) = z so now what? Im trying to find identity to get rid xy but I dont know how to do it ..

Where are you getting xz+z /z + zxy?

Be careful with your distribution. Once you distribute everything correctly, you'll see a common factor.

EDIT: Just realized you're using "/" as a "not" symbol. I suggest you use an apostrophe after the variable instead of a / before it.

 

Where are you getting xz+z /z + zxy?

Be careful with your distribution. Once you distribute everything correctly, you'll see a common factor.

EDIT: Just realized you're using "/" as a "not" symbol. I suggest you use an apostrophe after the variable instead of a / before it.

Sorry..
xz + z(x'+xy) = z <-- x' + xy = x' + y
xz + z ( x' + y ) = z
xz + zx' + zy = z < -- distributive
z ( x + x' + y ) = z <- x + x' = 1
z ( 1 + y ) = z <-- Im stuck! what can I do ?

 

z ( 1 + y ) = z <-- Im stuck! what can I do ?

what is any variable ORed with 1?

 

x' + xy = x' + y

I've never seen this identity, if that was taught as one, great, if not, I think you need to expand this out...

 

well in my notes that my teacher gave me says C' +CB = C+B

 

well in my notes that my teacher gave me says C' +CB = C+B

Well, then you're good. I've never seen given like that, but since your teacher gave that, use it!

 

but Im still stuck because you can see I carry out
xz + z ( x' + y ) = z
xz + zx' + zy = z < -- I expanded everything, and then the common factor is z and I got..
z ( x + x' + y ) = z <- since I have another x + x' = 1 according to my teacher A + A' = 1
z ( 1 + y ) = z <-- now Im not sure, I found another A + 1 = 1 ? is that can work 1 + y what do you think?

 

If anything is "OR"ed with a 1, the outcome is 1. So z (1 + y) is the same as z (1). What is Z AND 1?

 

If anything is "OR"ed with a 1, the outcome is 1. So z (1 + y) is the same as z (1). What is Z AND 1?

z (1 ) = z is that the way I should do it ?

 

z (1 ) = z is that the way I should do it ?

It works, doesn't it?

 

z (1 ) = z is that the way I should do it ?

...sounds like a logical progression

 

Now I have another that Im stuck sorry
x'+y' + xyz'= x'+y'+ z'
No idea how to do it .. the only think I can find x' y' could be xy'' ( xy are not) but there are not common factors or something in this equation that can lead me to do something

 

Now I have another that Im stuck sorry
x'+y' + xyz'= x'+y'+ z'
No idea how to do it .. the only think I can find x' y' could be xy'' ( xy are not) but there are not common factors or something in this equation that can lead me to do something

Remember the rule A' + AB' is A' + B'. If you rearrange a bit, you will see that this can help you twice.

Matt

 

Remember the rule A' + AB' is A' + B'. If you rearrange a bit, you will see that this can help you twice.

Matt

Sorry I dont know that rule, but I have this one with only one not .. A + A'B, but iff I have the whole term together because is XYZ' and the rule says A' + AB' ? is that applies?

 

Sorry I dont a rule like that, I have this one but looks like but I only have one not .. A + A'B in this case if I use your rule but if also applies if I have the whole term together because is XYZ' and the rule says A' + AB' ?

Hmm, I was thinking it worked with two NOTs just as it would with only 1. I could be wrong though. I haven't done this in a while

Perhaps I should let someone else jump in here, just in case my memory is failing me....

 

No problem I would wait lol I really need help :/

 

Remember the rule A' + AB' is A' + B'. If you rearrange a bit, you will see that this can help you twice.

Matt

Let:
A = x
B = y'

EDIT: formatted, per WBahn's advice...

x' + xy = x' + y
^    ^^   ^    ^
|    ||   |    |
A'   A B' A'   B'

 

Note: periods are for formatting, it didn't like spaces

Try putting it in a CODE block. You probably will still see it as proportional spacing while you are typing, but it will show up as monospaced when posted (or previewed).

 

A + A'B = A + B

This is confusing because it isn't an identity. It is valid, but you only see why when you go through the steps...
A + A'B : Expand A
A + AB + A'B : Factor B'
A + B(A + A') : Using Identity A + A' = 1
A + B(1) : Using identity A(1) = A
=> A + A'B = A + B

So, if we were to use another form of B, namely B':
A + A'B' : Expand A
A + AB' + A'B' : Factor B'
A + B'(A + A') : Using Identity A + A' = 1
A + B'(1) : Using identity A(1) = A
A + B'
=> A + A'B' = A + B'