Solve 23(11/12-7/8+3/4-13/24)*(-4*8)

$$23 ( \frac { 11 } { 12 } - \frac { 7 } { 8 } + \frac { 3 } { 4 } - \frac { 13 } { 24 } ) \times ( - 4 \cdot 8 )$$

Evaluate

$$-184$$

Solution Steps

Least common multiple of 12 and 8 is 24. Convert \frac{11}{12} and \frac{7}{8} to fractions with denominator 24.

$$23\left(\frac{22}{24}-\frac{21}{24}+\frac{3}{4}-\frac{13}{24}\right)\left(-4\right)\times 8$$

Since \frac{22}{24} and \frac{21}{24} have the same denominator, subtract them by subtracting their numerators.

$$23\left(\frac{22-21}{24}+\frac{3}{4}-\frac{13}{24}\right)\left(-4\right)\times 8$$

Subtract 21 from 22 to get 1.

$$23\left(\frac{1}{24}+\frac{3}{4}-\frac{13}{24}\right)\left(-4\right)\times 8$$

Least common multiple of 24 and 4 is 24. Convert \frac{1}{24} and \frac{3}{4} to fractions with denominator 24.

$$23\left(\frac{1}{24}+\frac{18}{24}-\frac{13}{24}\right)\left(-4\right)\times 8$$

Since \frac{1}{24} and \frac{18}{24} have the same denominator, add them by adding their numerators.

$$23\left(\frac{1+18}{24}-\frac{13}{24}\right)\left(-4\right)\times 8$$

Add 1 and 18 to get 19.

$$23\left(\frac{19}{24}-\frac{13}{24}\right)\left(-4\right)\times 8$$

Since \frac{19}{24} and \frac{13}{24} have the same denominator, subtract them by subtracting their numerators.

$$23\times \frac{19-13}{24}\left(-4\right)\times 8$$

Subtract 13 from 19 to get 6.

$$23\times \frac{6}{24}\left(-4\right)\times 8$$

Reduce the fraction \frac{6}{24} to lowest terms by extracting and canceling out 6.

$$23\times \frac{1}{4}\left(-4\right)\times 8$$

Multiply 23 and \frac{1}{4} to get \frac{23}{4}.

$$\frac{23}{4}\left(-4\right)\times 8$$

Express \frac{23}{4}\left(-4\right) as a single fraction.

$$\frac{23\left(-4\right)}{4}\times 8$$

Multiply 23 and -4 to get -92.

$$\frac{-92}{4}\times 8$$

Divide -92 by 4 to get -23.

$$-23\times 8$$

Multiply -23 and 8 to get -184.

$$-184$$

Factor

$$-184$$