Solve 1/3+5/2+20/7-(13/14+frac{1}{2})-frac{10}{7}+frac{11}{6}-(frac{1}{2})

$$\frac { 1 } { 3 } + \frac { 5 } { 2 } + \frac { 20 } { 7 } - ( \frac { 13 } { 14 } + \frac { 1 } { 2 } ) - \frac { 10 } { 7 } + \frac { 11 } { 6 } - ( \frac { 1 } { 2 } )$$

Evaluate

$$\frac{25}{6}\approx 4.166666667$$

Solution Steps

Least common multiple of 3 and 2 is 6. Convert \frac{1}{3} and \frac{5}{2} to fractions with denominator 6.

$$\frac{2}{6}+\frac{15}{6}+\frac{20}{7}-\left(\frac{13}{14}+\frac{1}{2}\right)-\frac{10}{7}+\frac{11}{6}-\frac{1}{2}$$

Since \frac{2}{6} and \frac{15}{6} have the same denominator, add them by adding their numerators.

$$\frac{2+15}{6}+\frac{20}{7}-\left(\frac{13}{14}+\frac{1}{2}\right)-\frac{10}{7}+\frac{11}{6}-\frac{1}{2}$$

Add 2 and 15 to get 17.

$$\frac{17}{6}+\frac{20}{7}-\left(\frac{13}{14}+\frac{1}{2}\right)-\frac{10}{7}+\frac{11}{6}-\frac{1}{2}$$

Least common multiple of 6 and 7 is 42. Convert \frac{17}{6} and \frac{20}{7} to fractions with denominator 42.

$$\frac{119}{42}+\frac{120}{42}-\left(\frac{13}{14}+\frac{1}{2}\right)-\frac{10}{7}+\frac{11}{6}-\frac{1}{2}$$

Since \frac{119}{42} and \frac{120}{42} have the same denominator, add them by adding their numerators.

$$\frac{119+120}{42}-\left(\frac{13}{14}+\frac{1}{2}\right)-\frac{10}{7}+\frac{11}{6}-\frac{1}{2}$$

Add 119 and 120 to get 239.

$$\frac{239}{42}-\left(\frac{13}{14}+\frac{1}{2}\right)-\frac{10}{7}+\frac{11}{6}-\frac{1}{2}$$

Least common multiple of 14 and 2 is 14. Convert \frac{13}{14} and \frac{1}{2} to fractions with denominator 14.

$$\frac{239}{42}-\left(\frac{13}{14}+\frac{7}{14}\right)-\frac{10}{7}+\frac{11}{6}-\frac{1}{2}$$

Since \frac{13}{14} and \frac{7}{14} have the same denominator, add them by adding their numerators.

$$\frac{239}{42}-\frac{13+7}{14}-\frac{10}{7}+\frac{11}{6}-\frac{1}{2}$$

Add 13 and 7 to get 20.

$$\frac{239}{42}-\frac{20}{14}-\frac{10}{7}+\frac{11}{6}-\frac{1}{2}$$

Reduce the fraction \frac{20}{14} to lowest terms by extracting and canceling out 2.

$$\frac{239}{42}-\frac{10}{7}-\frac{10}{7}+\frac{11}{6}-\frac{1}{2}$$

Least common multiple of 42 and 7 is 42. Convert \frac{239}{42} and \frac{10}{7} to fractions with denominator 42.

$$\frac{239}{42}-\frac{60}{42}-\frac{10}{7}+\frac{11}{6}-\frac{1}{2}$$

Since \frac{239}{42} and \frac{60}{42} have the same denominator, subtract them by subtracting their numerators.

$$\frac{239-60}{42}-\frac{10}{7}+\frac{11}{6}-\frac{1}{2}$$

Subtract 60 from 239 to get 179.

$$\frac{179}{42}-\frac{10}{7}+\frac{11}{6}-\frac{1}{2}$$

Least common multiple of 42 and 7 is 42. Convert \frac{179}{42} and \frac{10}{7} to fractions with denominator 42.

$$\frac{179}{42}-\frac{60}{42}+\frac{11}{6}-\frac{1}{2}$$

Since \frac{179}{42} and \frac{60}{42} have the same denominator, subtract them by subtracting their numerators.

$$\frac{179-60}{42}+\frac{11}{6}-\frac{1}{2}$$

Subtract 60 from 179 to get 119.

$$\frac{119}{42}+\frac{11}{6}-\frac{1}{2}$$

Reduce the fraction \frac{119}{42} to lowest terms by extracting and canceling out 7.

$$\frac{17}{6}+\frac{11}{6}-\frac{1}{2}$$

Since \frac{17}{6} and \frac{11}{6} have the same denominator, add them by adding their numerators.

$$\frac{17+11}{6}-\frac{1}{2}$$

Add 17 and 11 to get 28.

$$\frac{28}{6}-\frac{1}{2}$$

Reduce the fraction \frac{28}{6} to lowest terms by extracting and canceling out 2.

$$\frac{14}{3}-\frac{1}{2}$$

Least common multiple of 3 and 2 is 6. Convert \frac{14}{3} and \frac{1}{2} to fractions with denominator 6.

$$\frac{28}{6}-\frac{3}{6}$$

Since \frac{28}{6} and \frac{3}{6} have the same denominator, subtract them by subtracting their numerators.

$$\frac{28-3}{6}$$

Subtract 3 from 28 to get 25.

$$\frac{25}{6}$$

Factor

$$\frac{5 ^ {2}}{2 \cdot 3} = 4\frac{1}{6} = 4.166666666666667$$