Solve 21/3+(75/6-37/8*2)

$$2 \frac { 1 } { 3 } + ( 7 \frac { 5 } { 6 } - 3 \frac { 7 } { 8 } \times 2 )$$

Evaluate

$$\frac{29}{12}\approx 2.416666667$$

Solution Steps

Multiply 2 and 3 to get 6.

$$\frac{6+1}{3}+\frac{7\times 6+5}{6}-\frac{3\times 8+7}{8}\times 2$$

Add 6 and 1 to get 7.

$$\frac{7}{3}+\frac{7\times 6+5}{6}-\frac{3\times 8+7}{8}\times 2$$

Multiply 7 and 6 to get 42.

$$\frac{7}{3}+\frac{42+5}{6}-\frac{3\times 8+7}{8}\times 2$$

Add 42 and 5 to get 47.

$$\frac{7}{3}+\frac{47}{6}-\frac{3\times 8+7}{8}\times 2$$

Multiply 3 and 8 to get 24.

$$\frac{7}{3}+\frac{47}{6}-\frac{24+7}{8}\times 2$$

Add 24 and 7 to get 31.

$$\frac{7}{3}+\frac{47}{6}-\frac{31}{8}\times 2$$

Express \frac{31}{8}\times 2 as a single fraction.

$$\frac{7}{3}+\frac{47}{6}-\frac{31\times 2}{8}$$

Multiply 31 and 2 to get 62.

$$\frac{7}{3}+\frac{47}{6}-\frac{62}{8}$$

Reduce the fraction \frac{62}{8} to lowest terms by extracting and canceling out 2.

$$\frac{7}{3}+\frac{47}{6}-\frac{31}{4}$$

Least common multiple of 6 and 4 is 12. Convert \frac{47}{6} and \frac{31}{4} to fractions with denominator 12.

$$\frac{7}{3}+\frac{94}{12}-\frac{93}{12}$$

Since \frac{94}{12} and \frac{93}{12} have the same denominator, subtract them by subtracting their numerators.

$$\frac{7}{3}+\frac{94-93}{12}$$

Subtract 93 from 94 to get 1.

$$\frac{7}{3}+\frac{1}{12}$$

Least common multiple of 3 and 12 is 12. Convert \frac{7}{3} and \frac{1}{12} to fractions with denominator 12.

$$\frac{28}{12}+\frac{1}{12}$$

Since \frac{28}{12} and \frac{1}{12} have the same denominator, add them by adding their numerators.

$$\frac{28+1}{12}$$

Add 28 and 1 to get 29.

$$\frac{29}{12}$$

Factor

$$\frac{29}{2 ^ {2} \cdot 3} = 2\frac{5}{12} = 2.4166666666666665$$