Solve left(22/24-20/24+21/24-23/24right)*(-48)

$$\left( \frac{ 22 }{ 24 } - \frac{ 20 }{ 24 } + \frac{ 21 }{ 24 } - \frac{ 23 }{ 24 } \right) \times (-48)$$

Evaluate

$$0$$

Solution Steps

Reduce the fraction \frac{22}{24} to lowest terms by extracting and canceling out 2.

$$\left(\frac{11}{12}-\frac{20}{24}+\frac{21}{24}-\frac{23}{24}\right)\left(-48\right)$$

Reduce the fraction \frac{20}{24} to lowest terms by extracting and canceling out 4.

$$\left(\frac{11}{12}-\frac{5}{6}+\frac{21}{24}-\frac{23}{24}\right)\left(-48\right)$$

Least common multiple of 12 and 6 is 12. Convert \frac{11}{12} and \frac{5}{6} to fractions with denominator 12.

$$\left(\frac{11}{12}-\frac{10}{12}+\frac{21}{24}-\frac{23}{24}\right)\left(-48\right)$$

Since \frac{11}{12} and \frac{10}{12} have the same denominator, subtract them by subtracting their numerators.

$$\left(\frac{11-10}{12}+\frac{21}{24}-\frac{23}{24}\right)\left(-48\right)$$

Subtract 10 from 11 to get 1.

$$\left(\frac{1}{12}+\frac{21}{24}-\frac{23}{24}\right)\left(-48\right)$$

Reduce the fraction \frac{21}{24} to lowest terms by extracting and canceling out 3.

$$\left(\frac{1}{12}+\frac{7}{8}-\frac{23}{24}\right)\left(-48\right)$$

Least common multiple of 12 and 8 is 24. Convert \frac{1}{12} and \frac{7}{8} to fractions with denominator 24.

$$\left(\frac{2}{24}+\frac{21}{24}-\frac{23}{24}\right)\left(-48\right)$$

Since \frac{2}{24} and \frac{21}{24} have the same denominator, add them by adding their numerators.

$$\left(\frac{2+21}{24}-\frac{23}{24}\right)\left(-48\right)$$

Add 2 and 21 to get 23.

$$\left(\frac{23}{24}-\frac{23}{24}\right)\left(-48\right)$$

Subtract \frac{23}{24} from \frac{23}{24} to get 0.

$$0\left(-48\right)$$

Multiply 0 and -48 to get 0.

$$0$$

Factor

$$0$$