Math Problem Solver

\[ 1\frac{1}{7} + 4\frac{5}{6} = \frac{251}{42} = 5.9\overline{761904} \]

The sum can be computed by rewriting each term over a common denominator, where the least common denominator of \(\frac{1}{1}\), \(\frac{1}{7}\), \(\frac{4}{1}\), and \(\frac{5}{6}\) is \(42\). \[ \begin{aligned} 1\frac{1}{7} + 4\frac{5}{6} &= \frac{1 \cdot 42}{1 \cdot 42} + \frac{1 \cdot 6}{7 \cdot 6} + \frac{4 \cdot 42}{1 \cdot 42} + \frac{5 \cdot 7}{6 \cdot 7} \\ &= \frac{42 + 6 + 168 + 35}{42} \\ &= \frac{251}{42} \end{aligned} \]