While simplifying the above boolean expression. I have found to be using 2 methods yielding 2 different results.
A) By multiplying brackets only
A: $(x+y)(x+7)(\bar{x}+2)$ Since $\mathrm{XX}=\mathrm{X}$ ( By Idempotent Law) and $Y \bar{Y}=0$ $$ \begin{aligned} \text { L.H.S. } & =(X+X \bar{Y}+Y X+0)(\bar{X}+Z) \\ & =(X+X \bar{Y}+Y X)(\bar{X}+Z) \\ & =(X \bar{X}+X Z+X \bar{Y} \bar{X}+X \bar{Y} Z+Y X \bar{X}+Y X Z) \quad \text { (Distribution Law) } \\ & =(X \bar{X}+X Z+X \bar{X} \bar{Y}+X \bar{Y} Z+Y X \bar{X}+X Y Z) \quad \text { ( Commutative Law) } \end{aligned} $$ Since $X \bar{X}=0$ $$ \begin{aligned} \text { L.H.S. } & =(0+X Z+0 \bar{Y}+X \bar{Y} Z+Y 0+X Y Z) \\ & =(X Z+X \bar{Y} Z+X Y Z) \\ & =(X Z+X Z(Y+\bar{Y})) \end{aligned} $$
B) By multiplying brackets and simplifying 1st bracket inside:
$\begin{aligned} & \text { Q-9 }(x+y)(x+\bar{y})(\bar{x}+z) \\ & \rightarrow(x \cdot x+x \bar{y}+y x+y \bar{y})(\bar{x}+z) \\ & \rightarrow(x \cdot x = x) \rightarrow \text { Idempotent low } \\ &(x \cdot \bar{x} = 0) \rightarrow \text { Complement } \\ & \rightarrow(x+x \bar{y}+y x+0)(\bar{x}+z) \\ & \Rightarrow(x(\bar{y}+y)+x)(\bar{x}+z) \\ & \Rightarrow(x(1)+x)(\bar{x}+z)[\bar{y}+y=y] \\ & \rightarrow(x)(\bar{x}+z) \\ & \rightarrow x \bar{x}+x z \\ & \rightarrow 1+x z\end{aligned}$