Which law is this expression X+ X’.Y=X+Y

Question. Name the law given and verify it using a truth table. X+ X’.Y=X+Y

My Answer. 
X   Y   X’  X’.Y    X+X’.Y  X+Y
0   0   1   0       0       0
0   1   1   1       1       1
1   0   0   0       1       1
1   1   0   0       1       1

Prove algebraically that X + X’Y = X + Y.
L.H.S. = X + X’Y
           = X.1 + X’Y        (X . 1 = X property of 0 and 1)
           = X(1 + Y) + X’Y   (1 + Y = 1 property of 0 and 1)
           = X + XY +  X’Y                                                             
           = X + Y(X + X’)
           = X + Y.1          (X + X’ =1 complementarity law)
           = X + Y            (Y . 1 = Y property of 0 and 1)
           = R.H.S.      Hence proved.

My teacher marked my answer wrong. And told me to find the correct answer. Friends tell me is it a complementary law or distributive law or Absorption law? If it is absorption kindly tell me how to prove RHS and LHS algebraically.

One way of looking at this is as a consequence of distributivity, where $P+QR\equiv (P+Q)(P+R)$. Then you'll have
$$\begin{align}
X+(X'Y) &\equiv (X+X')(X+Y)&\text{distributivity}\\
&\equiv T(X+Y)&\text{inverse}\\
&\equiv X+Y&\text{domination}
\end{align}$$

StackExchange Computer Science Q#24587, answer score: 10