In the frieze group F7, write x-3zxyz in the form xnymz k.

It is known that $F_7=\left\{x^n{y}^m{z}^k\parallel n\in Z,m=0,k=0 or k=1\right\}$ and $x$ is the translation of $R^2$, and it map $\left(u,v\right)$ to $\left(u+a,v\right)$ for some constant $a$.

Also, $y$ is the horizontal reflection of $R^2$, and it map $\left(u,v\right)$ to $\left(u,-v\right)$.

Also, $z$ is the vertical reflection of $R^2$, and it map $\left(u,v\right)$ to $\left(-u,v\right)$.

So, it can be calculated that both $yz$ and $zy$ map $\left(u,v\right)$ to $\left(-u,-v\right)$.

So, it results,

$yz=zy \dots \left(1\right)$

Also it can be concluded that both $xy$ and $yx$ map $\left(u,v\right)$ to $\left(u+a,v\right)$.

so, it results,

$xy=yx \dots \left(2\right)$