Solve (-13/4+7/8-7/12)*(-11/7)

$$( - 1 \frac { 3 } { 4 } + \frac { 7 } { 8 } - \frac { 7 } { 12 } ) \times ( - 1 \frac { 1 } { 7 } )$$

Evaluate

$$\frac{5}{3}\approx 1.666666667$$

Solution Steps

Multiply 1 and 4 to get 4.

$$\left(-\frac{4+3}{4}+\frac{7}{8}-\frac{7}{12}\right)\left(-\frac{1\times 7+1}{7}\right)$$

Add 4 and 3 to get 7.

$$\left(-\frac{7}{4}+\frac{7}{8}-\frac{7}{12}\right)\left(-\frac{1\times 7+1}{7}\right)$$

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

$$\left(-\frac{14}{8}+\frac{7}{8}-\frac{7}{12}\right)\left(-\frac{1\times 7+1}{7}\right)$$

Since -\frac{14}{8} and \frac{7}{8} have the same denominator, add them by adding their numerators.

$$\left(\frac{-14+7}{8}-\frac{7}{12}\right)\left(-\frac{1\times 7+1}{7}\right)$$

Add -14 and 7 to get -7.

$$\left(-\frac{7}{8}-\frac{7}{12}\right)\left(-\frac{1\times 7+1}{7}\right)$$

Least common multiple of 8 and 12 is 24. Convert -\frac{7}{8} and \frac{7}{12} to fractions with denominator 24.

$$\left(-\frac{21}{24}-\frac{14}{24}\right)\left(-\frac{1\times 7+1}{7}\right)$$

Since -\frac{21}{24} and \frac{14}{24} have the same denominator, subtract them by subtracting their numerators.

$$\frac{-21-14}{24}\left(-\frac{1\times 7+1}{7}\right)$$

Subtract 14 from -21 to get -35.

$$-\frac{35}{24}\left(-\frac{7+1}{7}\right)$$

Add 7 and 1 to get 8.

$$-\frac{35}{24}\left(-\frac{8}{7}\right)$$

Multiply -\frac{35}{24} times -\frac{8}{7} by multiplying numerator times numerator and denominator times denominator.

$$\frac{-35\left(-8\right)}{24\times 7}$$

Do the multiplications in the fraction \frac{-35\left(-8\right)}{24\times 7}.

$$\frac{280}{168}$$

Reduce the fraction \frac{280}{168} to lowest terms by extracting and canceling out 56.

$$\frac{5}{3}$$

Factor

$$\frac{5}{3} = 1\frac{2}{3} = 1.6666666666666667$$