4(Z-4)/6 + 3Z-(1/6)=3/6
\(\frac{4(Z-4)}{6} + 3Z-\frac{1}{6}=\frac{3}{6}\\ \frac{4(Z-4)}{6} + \frac{18Z}{6}-\frac{1}{6}=\frac{3}{6}\\ \frac{4(Z-4)+18Z-1}{6} =\frac{3}{6}\\ 4(Z-4)+18Z-1=3\\ 4Z-16+18Z-1=3\\ 22Z-17=3\\ 22Z=20\\ 11Z=10\\ Z=\frac{10}{11}\)