Proving x=z Using Boolean Algebra Properties

[sjaguar13]

I think I am missing part of my notes, or at least I don't understand them:

if x+y = y+z and xy = xz, then x=z

x = (y+z)x absorption (Don't really know where this is coming from)
x(y+z) Commutative
xy+xz Distributive
It stops here and starts again at:

yz+xz (I have no idea where this came from)
zy+zx Commutative
z(y+z) Distributive
(z+y)z Here is where it gets confusing. Where did the z+y come from?
z Proven previously

 

[e(ho0n3]

I think I am missing part of my notes, or at least I don't understand them:

if x+y = y+z and xy = xz, then x=z

I think the easiest way of showing that is by using truth tables.

x = (y+z)x absorption (Don't really know where this is coming from)

x + y = y + z --> x(x + y) = x(y + z). Note that the left hand side simplifies to x, hence the result above.

yz+xz (I have no idea where this came from)

x + y = y + z --> (x + y)z = (y + z)z. The left hand side equals what you have above. Note that the right hand side simplifies to z.

zy+zx Commutative
z(y+z) Distributive
(z+y)z Here is where it gets confusing. Where did the z+y come from?

Just rearrange the terms from the previous step, i.e. z(y + z) = (z + y)z.

It's not clear to me what is going on here. Are these the steps from an exercise in a book?

[edit]I just noticed that these are supposedly notes. I've never taken any notes in any maths. class (or most classes for that matter). I think it would do you wonders to understand what's happening here rather than deciphering what notes you're missing.[/edit]

 

[sjaguar13]

It was an example the teacher did on the board, but I couldn't copy it down fast enough before it was time to go, so some of the lines I just scribbled and I am pretty sure some of the stuff is missing.

 

[supundika]

The solution for this problem

I think I am missing part of my notes, or at least I don't understand them:

if x+y = y+z and xy = xz, then x=z

x = (y+z)x absorption (Don't really know where this is coming from)
x(y+z) Commutative
xy+xz Distributive
It stops here and starts again at:

yz+xz (I have no idea where this came from)
zy+zx Commutative
z(y+z) Distributive
(z+y)z Here is where it gets confusing. Where did the z+y come from?
z Proven previously

Here the first statement (x=z)it self is the place you all have missed. And it has the soution. Hence you might be starting this formula from the middle.
Becuse of that x=(y+z)x is your starting point.

xy+xz becomes yz+xz because (x=z;first statement) so you have put z instead of x for xy and you haven't chaned xz which is perfectly correct.

And zy+zx becomes (z+y)z because you can write zy+zx as (y+x)z and it can be written as (z+y)z just by replacing x with z (x=z;first statement)

If you have a doubt about this or if you have any problem regarding maths please mail me through the following e mail supundikadl@yahoo.co.uk