Solve (1/3+1/7+1/2)div(2/3-frac{1}{2})

$$( \frac { 1 } { 3 } + \frac { 1 } { 7 } + \frac { 1 } { 2 } ) \div ( \frac { 2 } { 3 } - \frac { 1 } { 2 } )$$

Evaluate

$$\frac{41}{7}\approx 5.857142857$$

Solution Steps

Least common multiple of 3 and 7 is 21. Convert \frac{1}{3} and \frac{1}{7} to fractions with denominator 21.

$$\frac{\frac{7}{21}+\frac{3}{21}+\frac{1}{2}}{\frac{2}{3}-\frac{1}{2}}$$

Since \frac{7}{21} and \frac{3}{21} have the same denominator, add them by adding their numerators.

$$\frac{\frac{7+3}{21}+\frac{1}{2}}{\frac{2}{3}-\frac{1}{2}}$$

Add 7 and 3 to get 10.

$$\frac{\frac{10}{21}+\frac{1}{2}}{\frac{2}{3}-\frac{1}{2}}$$

Least common multiple of 21 and 2 is 42. Convert \frac{10}{21} and \frac{1}{2} to fractions with denominator 42.

$$\frac{\frac{20}{42}+\frac{21}{42}}{\frac{2}{3}-\frac{1}{2}}$$

Since \frac{20}{42} and \frac{21}{42} have the same denominator, add them by adding their numerators.

$$\frac{\frac{20+21}{42}}{\frac{2}{3}-\frac{1}{2}}$$

Add 20 and 21 to get 41.

$$\frac{\frac{41}{42}}{\frac{2}{3}-\frac{1}{2}}$$

Least common multiple of 3 and 2 is 6. Convert \frac{2}{3} and \frac{1}{2} to fractions with denominator 6.

$$\frac{\frac{41}{42}}{\frac{4}{6}-\frac{3}{6}}$$

Since \frac{4}{6} and \frac{3}{6} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\frac{41}{42}}{\frac{4-3}{6}}$$

Subtract 3 from 4 to get 1.

$$\frac{\frac{41}{42}}{\frac{1}{6}}$$

Divide \frac{41}{42} by \frac{1}{6} by multiplying \frac{41}{42} by the reciprocal of \frac{1}{6}.

$$\frac{41}{42}\times 6$$

Express \frac{41}{42}\times 6 as a single fraction.

$$\frac{41\times 6}{42}$$

Multiply 41 and 6 to get 246.

$$\frac{246}{42}$$

Reduce the fraction \frac{246}{42} to lowest terms by extracting and canceling out 6.

$$\frac{41}{7}$$

Factor

$$\frac{41}{7} = 5\frac{6}{7} = 5.857142857142857$$