Solve 1+1/3+frac{7{3}}

$$1+ \frac{ 1 }{ 3+ \frac{ 7 }{ 3 } }$$

Evaluate

$$\frac{19}{16}=1.1875$$

Solution Steps

Convert 3 to fraction \frac{9}{3}.

$$1+\frac{1}{\frac{9}{3}+\frac{7}{3}}$$

Since \frac{9}{3} and \frac{7}{3} have the same denominator, add them by adding their numerators.

$$1+\frac{1}{\frac{9+7}{3}}$$

Add 9 and 7 to get 16.

$$1+\frac{1}{\frac{16}{3}}$$

Divide 1 by \frac{16}{3} by multiplying 1 by the reciprocal of \frac{16}{3}.

$$1+1\times \frac{3}{16}$$

Multiply 1 and \frac{3}{16} to get \frac{3}{16}.

$$1+\frac{3}{16}$$

Convert 1 to fraction \frac{16}{16}.

$$\frac{16}{16}+\frac{3}{16}$$

Since \frac{16}{16} and \frac{3}{16} have the same denominator, add them by adding their numerators.

$$\frac{16+3}{16}$$

Add 16 and 3 to get 19.

$$\frac{19}{16}$$

Factor

$$\frac{19}{2 ^ {4}} = 1\frac{3}{16} = 1.1875$$