how do i solve 1/2(6-z)=z

thanks @melody

Does the (6-z) belong on the top as indicated by pedmas OR does it belong on the bottom?  Either way it would be good if you used more brackets to make your meaning clear.  Thanks. 

its exactky as the way its shown its just that im working with equations that have variables and thats the way it was shown to me 

Ok then, I will  answer the question that you have asked.  

$$\begin{array}{rll}
\frac{1}{2}(6-z)&=&z\\
\frac{1}{2}(6-z)\times 2&=&z \times 2\\
6-z&=&2z\\
6-z+z&=&2z+z\\
6&=&3z\\
\frac{6}{3}&=&\frac{3z}{3}\\
2&=&z\\
z&=&2\\
\end{array}$$

You are welcome.