## Elementary Technical Mathematics

9$\frac{5}{8}$ in
Let the two lenghs given are 6$\frac{3}{4}$ in and 2$\frac{7}{8}$ in Total length of the shaft = 6$\frac{3}{4}$ in + 2$\frac{7}{8}$ in =$\frac{27}{4}$ +$\frac{23}{8}$ =$\frac{27}{4}\times$ $\frac{2}{2}$ +$\frac{23}{8}$ = $\frac{54}{8}$ +$\frac{23}{8}$ =$\frac{77}{8}$ =9$\frac{5}{8}$ in