Solve (-2)^3*{[(7/8+4/15-1):(1/6-1-1/12+frac{1}{3}-frac{5}{6})+(-frac{1}{2}):(-frac{1}{2})]+(frac{3}{5}-frac{5}{8}+frac{3}{10})}

$$( - 2 ) ^ { 3 } \cdot \{ [ ( \frac { 7 } { 8 } + \frac { 4 } { 15 } - 1 ) : ( \frac { 1 } { 6 } - 1 - \frac { 1 } { 12 } + \frac { 1 } { 3 } - \frac { 5 } { 6 } ) + ( - \frac { 1 } { 2 } ) : ( - \frac { 1 } { 2 } ) ] + ( \frac { 3 } { 5 } - \frac { 5 } { 8 } + \frac { 3 } { 10 } ) \}$$

Evaluate

$$-\frac{47}{5}=-9.4$$

Solution Steps

Divide -\frac{1}{2} by -\frac{1}{2} to get 1.

$$\left(-2\right)^{3}\left(\frac{\frac{7}{8}+\frac{4}{15}-1}{\frac{1}{6}-1-\frac{1}{12}+\frac{1}{3}-\frac{5}{6}}+1+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Calculate -2 to the power of 3 and get -8.

$$-8\left(\frac{\frac{7}{8}+\frac{4}{15}-1}{\frac{1}{6}-1-\frac{1}{12}+\frac{1}{3}-\frac{5}{6}}+1+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Least common multiple of 8 and 15 is 120. Convert \frac{7}{8} and \frac{4}{15} to fractions with denominator 120.

$$-8\left(\frac{\frac{105}{120}+\frac{32}{120}-1}{\frac{1}{6}-1-\frac{1}{12}+\frac{1}{3}-\frac{5}{6}}+1+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Since \frac{105}{120} and \frac{32}{120} have the same denominator, add them by adding their numerators.

$$-8\left(\frac{\frac{105+32}{120}-1}{\frac{1}{6}-1-\frac{1}{12}+\frac{1}{3}-\frac{5}{6}}+1+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Add 105 and 32 to get 137.

$$-8\left(\frac{\frac{137}{120}-1}{\frac{1}{6}-1-\frac{1}{12}+\frac{1}{3}-\frac{5}{6}}+1+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Convert 1 to fraction \frac{120}{120}.

$$-8\left(\frac{\frac{137}{120}-\frac{120}{120}}{\frac{1}{6}-1-\frac{1}{12}+\frac{1}{3}-\frac{5}{6}}+1+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Since \frac{137}{120} and \frac{120}{120} have the same denominator, subtract them by subtracting their numerators.

$$-8\left(\frac{\frac{137-120}{120}}{\frac{1}{6}-1-\frac{1}{12}+\frac{1}{3}-\frac{5}{6}}+1+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Subtract 120 from 137 to get 17.

$$-8\left(\frac{\frac{17}{120}}{\frac{1}{6}-1-\frac{1}{12}+\frac{1}{3}-\frac{5}{6}}+1+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Convert 1 to fraction \frac{6}{6}.

$$-8\left(\frac{\frac{17}{120}}{\frac{1}{6}-\frac{6}{6}-\frac{1}{12}+\frac{1}{3}-\frac{5}{6}}+1+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Since \frac{1}{6} and \frac{6}{6} have the same denominator, subtract them by subtracting their numerators.

$$-8\left(\frac{\frac{17}{120}}{\frac{1-6}{6}-\frac{1}{12}+\frac{1}{3}-\frac{5}{6}}+1+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Subtract 6 from 1 to get -5.

$$-8\left(\frac{\frac{17}{120}}{-\frac{5}{6}-\frac{1}{12}+\frac{1}{3}-\frac{5}{6}}+1+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Least common multiple of 6 and 12 is 12. Convert -\frac{5}{6} and \frac{1}{12} to fractions with denominator 12.

$$-8\left(\frac{\frac{17}{120}}{-\frac{10}{12}-\frac{1}{12}+\frac{1}{3}-\frac{5}{6}}+1+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Since -\frac{10}{12} and \frac{1}{12} have the same denominator, subtract them by subtracting their numerators.

$$-8\left(\frac{\frac{17}{120}}{\frac{-10-1}{12}+\frac{1}{3}-\frac{5}{6}}+1+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Subtract 1 from -10 to get -11.

$$-8\left(\frac{\frac{17}{120}}{-\frac{11}{12}+\frac{1}{3}-\frac{5}{6}}+1+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Least common multiple of 12 and 3 is 12. Convert -\frac{11}{12} and \frac{1}{3} to fractions with denominator 12.

$$-8\left(\frac{\frac{17}{120}}{-\frac{11}{12}+\frac{4}{12}-\frac{5}{6}}+1+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Since -\frac{11}{12} and \frac{4}{12} have the same denominator, add them by adding their numerators.

$$-8\left(\frac{\frac{17}{120}}{\frac{-11+4}{12}-\frac{5}{6}}+1+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Add -11 and 4 to get -7.

$$-8\left(\frac{\frac{17}{120}}{-\frac{7}{12}-\frac{5}{6}}+1+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Least common multiple of 12 and 6 is 12. Convert -\frac{7}{12} and \frac{5}{6} to fractions with denominator 12.

$$-8\left(\frac{\frac{17}{120}}{-\frac{7}{12}-\frac{10}{12}}+1+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Since -\frac{7}{12} and \frac{10}{12} have the same denominator, subtract them by subtracting their numerators.

$$-8\left(\frac{\frac{17}{120}}{\frac{-7-10}{12}}+1+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Subtract 10 from -7 to get -17.

$$-8\left(\frac{\frac{17}{120}}{-\frac{17}{12}}+1+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Divide \frac{17}{120} by -\frac{17}{12} by multiplying \frac{17}{120} by the reciprocal of -\frac{17}{12}.

$$-8\left(\frac{17}{120}\left(-\frac{12}{17}\right)+1+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Multiply \frac{17}{120} times -\frac{12}{17} by multiplying numerator times numerator and denominator times denominator.

$$-8\left(\frac{17\left(-12\right)}{120\times 17}+1+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Cancel out 17 in both numerator and denominator.

$$-8\left(\frac{-12}{120}+1+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Reduce the fraction \frac{-12}{120} to lowest terms by extracting and canceling out 12.

$$-8\left(-\frac{1}{10}+1+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Convert 1 to fraction \frac{10}{10}.

$$-8\left(-\frac{1}{10}+\frac{10}{10}+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Since -\frac{1}{10} and \frac{10}{10} have the same denominator, add them by adding their numerators.

$$-8\left(\frac{-1+10}{10}+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Add -1 and 10 to get 9.

$$-8\left(\frac{9}{10}+\frac{3}{5}-\frac{5}{8}+\frac{3}{10}\right)$$

Least common multiple of 10 and 5 is 10. Convert \frac{9}{10} and \frac{3}{5} to fractions with denominator 10.

$$-8\left(\frac{9}{10}+\frac{6}{10}-\frac{5}{8}+\frac{3}{10}\right)$$

Since \frac{9}{10} and \frac{6}{10} have the same denominator, add them by adding their numerators.

$$-8\left(\frac{9+6}{10}-\frac{5}{8}+\frac{3}{10}\right)$$

Add 9 and 6 to get 15.

$$-8\left(\frac{15}{10}-\frac{5}{8}+\frac{3}{10}\right)$$

Reduce the fraction \frac{15}{10} to lowest terms by extracting and canceling out 5.

$$-8\left(\frac{3}{2}-\frac{5}{8}+\frac{3}{10}\right)$$

Least common multiple of 2 and 8 is 8. Convert \frac{3}{2} and \frac{5}{8} to fractions with denominator 8.

$$-8\left(\frac{12}{8}-\frac{5}{8}+\frac{3}{10}\right)$$

Since \frac{12}{8} and \frac{5}{8} have the same denominator, subtract them by subtracting their numerators.

$$-8\left(\frac{12-5}{8}+\frac{3}{10}\right)$$

Subtract 5 from 12 to get 7.

$$-8\left(\frac{7}{8}+\frac{3}{10}\right)$$

Least common multiple of 8 and 10 is 40. Convert \frac{7}{8} and \frac{3}{10} to fractions with denominator 40.

$$-8\left(\frac{35}{40}+\frac{12}{40}\right)$$

Since \frac{35}{40} and \frac{12}{40} have the same denominator, add them by adding their numerators.

$$-8\times \frac{35+12}{40}$$

Add 35 and 12 to get 47.

$$-8\times \frac{47}{40}$$

Express -8\times \frac{47}{40} as a single fraction.

$$\frac{-8\times 47}{40}$$

Multiply -8 and 47 to get -376.

$$\frac{-376}{40}$$

Reduce the fraction \frac{-376}{40} to lowest terms by extracting and canceling out 8.

$$-\frac{47}{5}$$

Factor

$$-\frac{47}{5} = -9\frac{2}{5} = -9.4$$