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Boolean algebra Simplify xy+x'yz'+yz
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factor out y from first two terms
y(x+x'z')+yz Just like that?
I think you can factor out the y from all three terms then you have y(x+x'z'+z) but we can re-write that y((x+z)+x'z') then use the identity (x+z) = (x'z')' to get y((x'z')'+x'z') which is, of course, just y.
nice :) @genson0 try below also : y(x+x'z')+yz y(x+z') + yz y(x + z' + z) y(x + 1) y(1) y
Thank you both @ganeshie8 and @Valpey ,, I appreciate the help. I wish I could give you both medals.
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