\(\begin{aligned} & & y & =\frac{1}{2}\left(z+\frac{1}{z}\right)\\ & \implies & y & =\frac{1}{2}z+\frac{1}{2}\cdot\frac{1}{z}\\ & \implies & y & =\frac{1}{2}z+\frac{1}{2z}\\ & \implies & 2z\cdot y & =2z\cdot\left(\frac{1}{2}z+\frac{1}{2z}\right)\\ & \implies & 2z\cdot y & =2z\cdot\frac{1}{2}z+2z\cdot\frac{1}{2z}\\ & \implies & 2z\cdot y & =\frac{2z\cdot1\cdot}{2}+\frac{2z\cdot1}{2z}\\ & \implies & 2z\cdot y & =\frac{z}{2}+1\\ & \implies & 2z\cdot y-\frac{z}{2} & =1\\ & \implies & \left(2y-\frac{1}{2}\right)z & =1\\ & \implies & z & =\frac{1}{2y-\frac{1}{2}} \end{aligned}\)