K-Map to solve algebraic reduction - boolean

Question

I need help verifying an algebraic expression using K-Map.
The expression I'm posting was actually done by my professor, but for practice purpose I wanted to use the K-Map to verify that the answer is correct.
-> X • Y + X' • Y • Z' + Y • Z =
-> X • Y •(Z + Z') + X' • Y • Z' + Y • Z =
-> X•Y•Z + X•Y•Z' + X'•Y•Z' + Y• Z =
-> X•Y•Z + Y•Z' • (X + X') + Y•Z =
-> X•Y•Z + Y•Z' + Y•Z =
-> X•Y•Z + Y•(Z' + Z) =
-> X•Y•Z + Y=
-> Y•(X•Z + 1)= Y
When I tried my K-Map solution to prove the answer is correct, I came up with this
YZ' + X'Y
I'm not sure if my answer is correct or not. If it's not, I need someone to show me how to make it correct, so that it matches the solution I've posted. I appreciate the help.

Answer 1

Remember that a 1 goes into the K-map if and only if the expression doesn't change and match with either two 0s or two 1s. The expression does not exist when it changes or has a 0 and 1 together.
So this is my K-Map and what I did.
Kmap
My answer came up as Y, which also matches your professor's answer.
(Edit: I forgot to put the pluses in the equation, but I hope you get the idea)

Answer 2

The deduction and answer given by your professor is correct (every step is). Therefore your answer, being different, is not correct. Take for example the case X = Y = Z = 1. The original expression evaluates as follows
X • Y + X' • Y • Z' + Y • Z = 1 • 1 + 0 • 1 • 0 + 1 • 1
= 1 + 0 + 1
= 1
However,
Y • Z' + X' • Y = 1 • 0 + 0 • 1
= 0 + 0
= 0
which is different from the original expression and also from its equivalent Y.

Answer 3

You can break up the given function in the form of SOP (DNF) into separate terms and find them in the K-map. It is really like a crisscross.
Having the function:
f(x,y,z) = x·y + ¬x·y·¬z + y·z
where:
x·y ~ cells, where the x is true and the y is true; marked red;
¬x·y·¬z ~ cells, where the x and z are false and the y is true; marked blue; because it is using all the given variables, this term is also an minterm marking out only a single cell;
y·z ~ cells, where both y and z are true; marked green.
You can see, that the marked cells can be grouped into bigger 2^i sized bubble, a 2^2=4 one.
And you should use that instead of having the three products in the original function, because the less the variables and operators the easier to work with the expression.
Circling the largest possible group, you get the same result as your teacher.
f(x,y,z) = x·y + ¬x·y·¬z + y·z = y
As you can see now the expression logical value is in 50 percent of cases evaluated as true and in other 50 percent false, because the value depends only on the value of variable y.