Solve 21/3+(33/5div11/8)

$$2 \frac { 1 } { 3 } + ( 3 \frac { 3 } { 5 } \div 1 \frac { 1 } { 8 } )$$

Evaluate

$$\frac{83}{15}\approx 5.533333333$$

Solution Steps

Multiply 2 and 3 to get 6.

$$\frac{6+1}{3}+\frac{\frac{3\times 5+3}{5}}{\frac{1\times 8+1}{8}}$$

Add 6 and 1 to get 7.

$$\frac{7}{3}+\frac{\frac{3\times 5+3}{5}}{\frac{1\times 8+1}{8}}$$

Divide \frac{3\times 5+3}{5} by \frac{1\times 8+1}{8} by multiplying \frac{3\times 5+3}{5} by the reciprocal of \frac{1\times 8+1}{8}.

$$\frac{7}{3}+\frac{\left(3\times 5+3\right)\times 8}{5\left(1\times 8+1\right)}$$

Multiply 3 and 5 to get 15.

$$\frac{7}{3}+\frac{\left(15+3\right)\times 8}{5\left(1\times 8+1\right)}$$

Add 15 and 3 to get 18.

$$\frac{7}{3}+\frac{18\times 8}{5\left(1\times 8+1\right)}$$

Multiply 18 and 8 to get 144.

$$\frac{7}{3}+\frac{144}{5\left(1\times 8+1\right)}$$

Multiply 1 and 8 to get 8.

$$\frac{7}{3}+\frac{144}{5\left(8+1\right)}$$

Add 8 and 1 to get 9.

$$\frac{7}{3}+\frac{144}{5\times 9}$$

Multiply 5 and 9 to get 45.

$$\frac{7}{3}+\frac{144}{45}$$

Reduce the fraction \frac{144}{45} to lowest terms by extracting and canceling out 9.

$$\frac{7}{3}+\frac{16}{5}$$

Least common multiple of 3 and 5 is 15. Convert \frac{7}{3} and \frac{16}{5} to fractions with denominator 15.

$$\frac{35}{15}+\frac{48}{15}$$

Since \frac{35}{15} and \frac{48}{15} have the same denominator, add them by adding their numerators.

$$\frac{35+48}{15}$$

Add 35 and 48 to get 83.

$$\frac{83}{15}$$

Factor

$$\frac{83}{3 \cdot 5} = 5\frac{8}{15} = 5.533333333333333$$